


■f.-.f •■' '--.'.':• '-'^'•y>;t^'-y<f-'^P'nCta;-y^>e4v-*''. ■ 




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COP^'RIGHT DEPOSIT. 



ENGINEERING 
THERMODYNAMICS 



BY 

JAMES AMBROSE MOYER, S. B., A. M. 

Professor of Mechanical Engineermg in The 
Fenfisylvania State College 

. AND 

JAMES PARK CALDERWOOD, M. E. 

Associate Professor of Mechanical Engineering in The 
Pennsylvania State College 



FIRST EDITION 

FIRST THOUSAND 



NEW YORK 

JOHN WILEY & SONS, Inc. 

London: CHAPMAN & HALL, Limited 

1915 






Copyright, 191 5, 

BY 

J. A. MOYER AND J. P. CALDERWOOD 



/i^^^-^ 



fc) 



stanhope iprcss 

F. H. GILSON COMPANY 
BOSTON, U.S.A. 



NOV I 1915 

©CI.A414330 



PREFACE 



For years there has been an important demand for a text- 
book on thermodynamics which would be brief and concise, but 
at the same time so clearly written as regards explanation that 
students of average ability in our large technical schools could 
read it without difhculty. A professor of thermodynamics wrote 
recently as follows: ''I like the idea of making the text largely 
self-explanatory. Too many books require the reading of several 
lines between every two Hnes of the text." This book has been 
prepared to meet this demand and in writing it the authors have 
kept in mind these requirements. Further, it has been the idea 
of the authors to make this book particularly suitable for use in 
the larger technical schools where it is possible to give special 
courses on the subjects of steam turbines, pneumatic machinery, 
internal combustion engines, refrigeration and pumping ma- 
chinery. Usually in the courses on these subjects the advanced 
and special theory of thermodynamics as it relates to each of 
them is taken up with completeness. At present there is too 
much duplication of subject matter in the large volumes on 
thermodynamics now available, which are made to include the 
descriptive matter and the applications of these special subjects. 

The authors are particularly indebted to Professor Roy B. Fehr 
of The Pennsylvania State College for invaluable assistance and 
criticisms in the preparation of this work. Acknowledgments 
are due also to President Ira N. Hollis of Worcester Polytechnic 
Institute; Professor Lionel S. Marks of Harvard University and 
Massachusetts Institute of Technology; Professor H. C. Ander- 
son of the University of Michigan; Dr. S. A. Moss of the 
General Electric Company; Dr. William Kent of Montclair, 
N. J.; Professor A. M. Greene of Rensselaer Polytechnic In- 



IV PREFACE 

stitute; Professor A. L. Westcott of University of Missouri; 
Professor A. A. Atkinson of Ohio University; and Professors 
J. A. Bursley and C. H. Fessenden of the University of Michigan, 
for assistance in various ways. 

A book on applied thermodynamics published privately by 
the late Professor H. W. Spangler and with the preparation of 
which one of the authors was intimately associated has been 
consulted freely in the preparation of the last chapters. This 
book, as a whole, includes many of Professor Spangler's ideas as 
regards subject matter to be included in a book of this kind. 

Acknowledgment of the services of Messrs. W. M. Sides and 
H. J. Hartranft of State College, Pa., is due for assistance in 
preparation and proof-reading. 

J. A. MOYER 

J. P. Calderwood 

State College, Pa., 
August, 1915. 



CONTENTS 



Page 
Chapter I. — Introduction to the Theory or Heat Engines i 

Historical; Definition, Scope, and Object of Thermodynamics; First and 
Second Laws of Thermodynamics; Units Employed: British Thermal 
Units, Density, Specific Volume, Mechanical Equivalent of Heat; Ther- 
mal Efficiency; Working Substances. 

Chapter II. — Properties of Perfect Gases 9 

Examples of Perfect Gases and Vapors; Laws of Perfect Gases: Boyle's 
Law, Charles' Law, Combination Law (PV = MRT); Absolute Tempera- 
ture and Pressure; Gas Thermometers; Heat and Its Effects : Heat Added 
= Increase of Internal Energy + Work Done; Internal or Intrinsic 
Energy; External Work; Specific Heats : Cj, and C^; Joule's Law; Rela- 
tion between Cp and Cv', Ratio of Specific Heats; Low Temperature Re- 
searches. 

Chapter III. — Expansion and Compression of Gases 29 

Graphical Diagrams; Indicator Diagrams; Isothermal Expansion and 
Compression; Adiabatic Expansion and Compression; Change of Internal 
Energy during Adiabatic Processes. 

Chapter IV. — Cycles of Heat Engines 44 

Camot's Cycle; Heat Engines and Refrigerating Machines; Efficiency of 
Camot's Cycle; Reversible Cycles; Camot's Principle; Perfection in a 
Heat Engine; Reversed Camot's Cycle; Regenerative Air Engines. 

Chapter V. — Properties of Steam 60 

Steam in Heat Engines; Process of Steam Formation; Relation of Tem- 
perature, Pressure and Volume in Saturated Steam; Heat of the Liquid; 
Latent Heat of Vaporization; External Work of Evaporation; Total Heat 
of Steam; Intemal Energy of Evaporation and of Steam; Steam Formed 
at Constant Volume; Wet and Dry Saturated Steam; Superheated Steam; 
Determination of Moisture in Steam; Throtthng or Superheating Calo- 
rimeters; Separating Calorimeters; Condensing or Barrel Calorimeters; 
Equivalent Evaporation; Factor of Evaporation, 

Chapter VI. — Practical x\pplications of Thermodynamics 88 

Refrigerating Machines or Heat Pumps; Systems in Use; Coefficients of 
Performance; Compressed Air; Indicator Diagrams from Compressors; 
Comparison of Actual and Ideal Compression Curves; Multi-stage Com- 
pressors. 



VI CONTENTS 

Page 
Chapter VII. — Entropy loi 

Pressure- volume Diagrams and their Uses; Heat Diagrams; Explanation 
of Entropy; Camot's Cycle on a Temperature-entropy Diagram; Tem- 
perature-entropy Diagrams for Steam; Isothermal and Adiabatic Lines 
for Steam; Saturation Line. 

Chapter VIII. — Practical Steam Expansions and Cycles 113 

Steam Engine Efficiency With and Without Expansion of the Steam; 
Quality of Steam during Adiabatic Expansion; MoUier Diagram; Graphical 
Determination of Quality of Steam by Throttling Calorimeter; Available 
Energy from Adiabatic Expansion; Rankine Cycle; EfiQciency of Various 
Prime Movers; Combined Indicator Diagrams; Hirn's Analysis of the 
Steam Engine. 

Chapter IX. — Flow of Fluids 163 

Flow of Air; Thermodynamic Formulas; Fliegner's Formulas; Experi- 
mental Data; Flow of Steam; Grashof's Formula; Napier's Formula; 
Flow through Nozzles; Restriction of Flow Due to Over or Under Expan- 
sion; Correction for Flow in Poorly Designed Nozzles. 



SYMBOLS 



A = area in square feet. 
B.t.u. = British thermal units (=778 ft. lbs.). 

Cy = specific heat at constant pressure in B.t.u. per pound per degree. 
Cv = specific heat at constant volume in B.t.u. per pound per degree. 
C = a general constant in equations of perfect gases. 
D = degrees of superheat. 
E = external work in B.t.u. per pound; also sometimes used to express efficiency, 

usually as a decimal. 
Ea = available energy in B.t.u. per pound. 
F = force in pounds. 
H = heat per pound in B.t.u.* 
^sup = total heat of superheated steam, B.t.u. per pound. 
Ih = total internal energy of steam (above 32° F.) in B.t.u. per pound. 
Il — internal energy of evaporation of steam in B.t.u. per pound. 
/ = reciprocal of mechanical equivalent of heat = yf g (use becoming obsolete) . 
K = specific heat in foot-pound units. 
L = latent heat of evaporation in B.t.u. per pound. 
M = mass (pounds). 

P = pressure in general or pressure in pounds per square foot. 
Q = quantity of heat in B.t.u. 
R = thermodynamic constant for gases; for air it is 53.3 (in foot-pound units 

per pound.) 
T = absolute temperature, in Fahr. degrees = 460 -|- t. 

V = volume in cubic feet, also specific volume and velocity in feet per second. 
W. E. = Warme Einheit = kilogram calorie. 
W = work done in foot-pounds. 
a = area in square inches. 
c = constant of integration. 
d = distance in feet. 

e = subscript to represent base of natural logarithms. 
g = acceleration due to gravity = 32.2 feet per second per second. 
h = heat of the hquid per pound in B.t.u. (above 32° F.). 
- k = a. constant, 
log = logarithm to base 10. 
loge = logarithm to natural base e (Naperian) . 

n = general exponent for V (volume) in equations of perfect gases, also some- 
times used for entropy of the Uquid in B.t.u. per degree of absolute tem- 
perature. 

* In steam tables it is total heat above 32° F. 
vii 



vm SYMBOLS 

p = pressure in pounds per square inch. 

q = sometimes used for heat of the liquid in B.t.u. per pound (above 32° F.). 

r = ratio of expansion (see page $3) > also sometimes used for latent heat of 

evaporation in B.t.u. per pound. 
/ = temperature in ordinary Fahr. degrees. 

V = specific volume, in cubic feet per pound (in some steam tables). 
w = weight per cubic foot = density. 
X = quality of steam expressed as a decimal. 

A = differential symbol. 

(^ 
y = ratio of specific heats -pr" 

dO 

<i> = total entropy -^• 

= entropy of the liquid in B.t.u. per pound per degree of absolute temperature. 



Engineering Thermodynamics 

CHAPTER I 
INTRODUCTION TO THE THEORY OF HEAT ENGINES 

The beginning of our knowledge of the theory of heat on a 
scientific basis is probably Black's doctrine of latent heat.* The 
important relations as to the equivalence of heat and work were 
not established until about the middle of the nineteenth century, 
long after the heat engines invented by Newcomen and Watt 
had found general use. The greatest development in the theory 
of heat as regards its use in engines started with the pubhca- 
tions of Carnot in 1824, when he showed that work is performed 
by a heat engine in the same proportion as the absolute tem- 
perature of the working fluid is changed from a higher to a lower 
degree. The next important development in heat theory was 
made by Joule who showed by his experiments in 1843 ^^^-t for 
every unit of heat there was an exact equivalent in mechanical 
work. Although these experiments and the conclusions from 
them were epoch-making, there was Httle attempt at accurate 
calculations for designing steam and other engines until Reg- 
nault in 1847 published his classical data on the properties of 
steam. In the next few years important contributions to our 
knowledge of what the ideal engine should be, particularly as 
regards its efficiency in the way of heat conversion, were made 
by Rankine, Clausius, and Lord Kelvin. It was undoubtedly 
Rankine's philosophical treatment of the subject, as presented 
in various books and papers of which he was the author, that 
gave steam engineers scientific methods for the development of 
new and improved designs. 

Thermodynamics is that branch of engineering science which 
deals with the interconversion of heat and work.f The object 

* Encyclopedia Britannica, iithed., vol. 4, page 18. 

t In general, thermodynamics may be defined as that branch of physical science 
which treats of the effects produced by heat. 



2 ENGINEERING THERMODYNAMICS 

of the study of thermodynamics is to consider in the light of the 
most recent investigations, questions relating to the influence 
on the efficiency of heat engines of increased pressure, of higher 
vacuums and expansions, of higher superheats (meaning also a 
greater range of expansion), of jacketing cylinders, of using the 
working medium in several cylinders one after the other; that 
is, compounding instead of expanding it only in a single cylinder, 
of putting receivers between the cylinders, and of reheating the 
working medium as it passes through them from one cylinder 
to the next. All these subjects require the most careful con- 
sideration by the engineer, even though theoretical considera- 
tions are not always directly applicable, and the final results 
which determine our engineering practice are largely determined 
by the theoretical analysis of carefully conducted experiments. 
The student should, therefore, keep in mind that although the 
actual conditions existing as regards the operation of our heat 
engines are relatively complex, the exact theory of their action is 
of great practical value, and he should, for this reason, give the 
study of the theory of heat engines his best attention. Because, 
also, of this complexity in the action of engines the study of 
the theory underlying their operation becomes all the more 
essential, as it must serve as a guide in deciding what conditions 
are most important for securing the highest efficiency. 

Another important service which the study of thermodynamics 
renders is that of showing us what maximum efficiency is attain- 
able for any engine operating under a given set of conditions. 
It often happens that the enthusiastic inventor presents data 
showing results which indicate an efficiency very much better 
than we can obtain with any of our present types of engines. 
In such cases it requires usually only a very little calculation, 
starting from the fundamental theory of thermodynamics, to 
show conclusively that the results claimed are absolutely im- 
possible. 

In many cases this study is also useful to the engineer in check- 
ing up his own work. For example, in calculating the results of 
a test made on a gas engine of the ordinary types, if it is found 



INTRODUCTION TO THE THEORY OF HEAT ENGINES 3 

that the efficiency as computed and reported is higher than 
50 per cent, then obviously our knowledge of thermodynamics, 
and of the maximum possible efficiency of such engines operat- 
ing under ordinary conditions, shows that the results are impos- 
sible and absurd. The ability to interpret correctly the results 
of experiments performed on all kinds of heat engines and simi- 
lar apparatus requires a thorough knowledge of the basic prin- 
ciples of thermodynamics. 

Laws of Thermodynamics. The heat engine in its simplest 
form does work by utilizing heat developed in a furnace or 
generated by the combustion of fuel within the engine itself. A 
part of the heat thus suppHed is spent in doing mechanical work 
so that it no longer exists in the form of heat energy, while the 
remainder is rejected by the engine, still in the form of heat. 
The heat action thus described depends on two fundamental 
laws of thermodynamics called generally the First and Second. 

First Law of Thermodynamics. The statement of the first 
of these is that the amount of heat which disappears in the 
action of the heat engine is proportional to the amount of me- 
chanical work performed by the engine. In other words this 
law is nothing more than a simple statement of the conserva- 
tion of energy as regards the equivalence of mechanical work 
and heat. With more exactness this first law may be stated 
as follows: 

A definite quantity of heat goes out of existence for every 
unit of mechanical work that is performed, and, conversely, 
when heat is developed by the performance of mechanical work, 
a definite quantity of heat comes into existence for every unit 
of work. 

Second Law of Thermodynamics. A self-acting machine can- 
not transmit heat from one body at a lower temperature to 
another body at a higher temperature unless aided by some ex- 
ternal agency; that is, " heat cannot pass from a cold body to 
a hot body by a purely self-acting process" (Clausius). This 
law really states as regards heat engines the limits to their pos- 
sible performance, which would be otherwise unlimited, if only 



4 ENGINEERING THERMODYNAMICS 

the "first law" is considered. It states further, as will be 
discovered in the discussion that follows, that no heat engine 
converts or can convert into work all of the heat supplied to 
it. A very large part of the heat suppHed is necessarily rejected 
by the engine in the form of unused heat. 

ENGINEERING UNITS 

Engineers in English-speaking countries use as the standard 
of measurements the units generally known as the foot-pound- 
second-system. In this system the unit of length is the foot, 
the unit of mass is the pound (equivalent to 0.4535 kilogram), 
and the unit of time is the second. On the same basis we use 
as derived units the square foot as the unit of area in all theo- 
retical calculations, unless it is expressly stated that the area 
is given in some other units, as for example in square inches. 
Practically all equations and formulas are stated with the volume 
given in cubic feet. Unit pressure, or what we call briefly 
** pressure," is the total applied force in pounds divided by the 
total area in square feet over which it is exerted, and is then 
expressed in terms of pounds per square foot. 

Specific Volume is the term applied to the volume in cubic 
feet of a pound of a substance. 

Density is the mass in pounds, of a cubic foot of a substance. 
It is, therefore, the reciprocal of the specific volume. 

For engineering purposes the ordinary unit of work is the 
foot-pound, which is the quantity of work performed by a force 
of one pound in moving through a distance of one foot. Quan- 
tities of heat are usually expressed in terms of the British ther- 
mal unit (B.t.u.). This unit is the quantity of heat required to 
raise the temperature of one pound of water one degree on the 
Fahrenheit scale. The corresponding unit of heat on the Centi- 
grade scale, used almost universally by physicists and chemists, 
is called the calorie (French and English), and Warme Einheit 
(German).* It is the quantity of heat required to raise the 

* Temperatures in Centigrade degrees are converted into Fahrenheit by mul- 
tiplying by I and adding 32. Kilogram-calories multiplied by 3.968 give the 



INTRODUCTION TO THE THEORY OF HEAT ENGINES 5 

temperature of one gram of water one degree on the Centigrade 
scale. 

To make the definition of the B.t.u. accurate it is necessary 
to state at what temperature the rise of 1° F. as stated is to occur, 
because the specific heat of water is sHghtly variable. This is 
also one of the reasons for some of the variations in the tables 
of the properties of steam and other vapors that we shall observe 
in our calculations. Some tables are based on the assumption 
that the B.t.u. is the amount of heat required to raise the tem- 
perature of water 1° F. at the condition of maximum density of 
water, that is, between 39° and 40° F. Other tables are based 
on the amount of heat required to raise the temperature 1° F. 
from 60 to 61 degrees. Still another table, which is the latest 
and is generally considered to be the most accurate, uses for the 
B.t.u. one one-hundred-and-eightieth (yjo) ^^ ^^^ amount of heat 
required to raise the temperature of water from 32° to 2i2°F. 
In other words, according to this last definition the B.t.u. is 
the average value of the amount of heat required to raise the 
temperature of one pound of water one degree between the con- 
ditions of freezing and boiling at atmospheric pressure. 

Mechanical Equivalent of Heat. It has already been stated 
that for every heat unit there is a corresponding exact equivalent 
which can be expressed in the mechanical units of work. The 
number of work units corresponding to i B.t.u. is called the 
mechanical equivalent of heat. This quantity was necessarily 
determined by experiments. The most reliable results show 
that the average value of the mechanical equivalent of heat 
is 778 foot-pounds. This means, in other words, that a given 
number of B.t.u. multipHed by 778 gives the number of foot- 
pounds of work corresponding to the amount of heat as ex- 
pressed by the thermal units. 

This value of the mechanical equivalent of heat is the one 
given by Rowland, Griffith, Osborne Reynolds and Morby. 

equivalent British thermal units (.B.t.u.), and kilogram-calories per kilogram 
times 1.8 give B.t.u. per pound. A "small" calorie, or gram-calorie, is one- 
thousandth as large as a kilogram-calorie. 



6 ENGINEERING THERMODYNAMICS 

The first determination of this factor was made by Joule. In 
1843 h^ obtained the value of 772 foot-pounds. In later experi- 
ments he found the value to lie between 774 and 775. A great 
many modern physicists believe that the most exact value is 
somewhere between 779 and 780 foot-pounds. The value stated 
above as given by Rowland and others, that is, 778 foot-pounds, 
is, however, the value most generally accepted and is invariably 
used in engineering calculations. 

THERMAL EFFICIENCY OF A HEAT ENGINE 

Heat converted into work 



Thermal efficiency = 



Heat suppHed to the engine 

Since only a part of the heat supplied to an engine can be con- 
verted into work, the above ratio is a fraction always less than 
unity. 

Since heat is not a form of matter, it must be transmitted, in 
a physical sense, by the use of some medium which is called the 
working substance. It is this working substance that takes in 

Spring 
Detent 




Fig. I. — "Ratchet and Pawl" Type of Heat Engine. 

and rejects heat. It may be in the form of a gas, a liquid or a 
soHd. In general, there is a considerable change of volume in 
a suitable working substance for heat engines as that substance 
does work in overcoming resistance. 



INTRODUCTION TO THE THEORY OF HEAT ENGINES 7 

As the simplest imaginable example of a heat engine in which 
the working substance is neither a gas nor a liquid, but a solid 
material, we may think of the working substance of the engine 
as being a long rod of brass arranged to operate as the pawl 
of a ratchet wheel (Fig. i) with teeth relatively close together. 
Now if the pawl is heated, it will elongate sufficiently to drive the 
tooth of the ratchet wheel with which it is in contact far enough 
forward to have the wheel held in its last position by a click or 
detent. When the pawl cools, as, for example, by pouring on 
cold water, it will contract and fall into the position where it 
will engage the next succeeding tooth on the wheel, which upon 
application of heat will again be driven forward and held in 
position, while the contracting of the pawl is again repeated. 
In fact, if the ratchet wheel referred to above is fastened at its 
center to a round shaft, its movement can be readily made to do 
work by raising the weight attached to a cord winding on the 
shaft. This heat engine has heat supplied to it at a high tem- 
perature. A small part of this heat is transformed into mechan- 
ical work in moving the ratchet wheel and with it the weight to 
be lifted, but by far the greatest part of the heat supplied is 
rejected in being absorbed by the water used for cooling and 
required to bring about the necessary contraction of the brass 
rod which is the working substance. This action as described 
here for the metaUic working substance is typical of the action 
of all heat engines. They must take in heat at a relatively 
high temperature and reject it at a comparatively low tempera- 
ture, and in the process they convert a small amount of heat 
supplied into mechanical work. This process is also somewhat 
similar to the performance of a water wheel which does work 
in bringing water from a higher to a lower level. In the case 
of the water wheel we say that the water has lost potential 
energy in dropping from a higher to a lower level. Similarly 
the working substance in heat engines loses heat energy, which 
is in fact (lost) potential energy as it drops from a higher to a 
lower temperature. 



8 ENGINEERING THERMODYNAMICS 

PROBLEMS 

1. One pound of fuel has a heating value of 14,500 B.t.u. How manj 
foot-pounds of work is it capable of producing, if all this heat could b( 
converted into work ? Ans. 11,281,000 ft. -lbs. 

2. An engine developed 15,560 ft. -lbs. of work. How much heat was 
required? Ans. 20 B.t.u. 

3. A stone weighing 2 lbs. is let fall from a height of 389 ft. into i lb, 
of water. How many degrees is the water raised? Ans. 1° F. 

4. A heat engine receives 100,000 B.t.u. of heat in the form of fuel 
and during the same period 30,000 B.t.u. are converted into work. What 
is the thermal efificiency of the engine? Ans. 30 per cent. 

5. A gas engine receives 20,000 B.t.u. of heat in the form of fuel, and 
during the same period 3,112,000 ft. -lbs. of work are developed. What is 
the thermal efhciency of the engine? Ans. 20 per cent. 

6. Show that kilogram-calories per kilogram X 1.8 give B.t.u. per pound. 

7. Convert —40° C. into degrees Fahrenheit. Ans. —40° F. 



CHAPTER II 



PROPERTIES OF PERFECT GASES 

In the study of thermodynamics we shall have to deal mostly 
with fluids in the condition of a perfect gas, or a vapor. When 
the word ^' gas " is used it refers to what is more properly 
called a perfect gas, which is a fluid remaining in the gaseous 
condition even when subjected to moderately high pressures 
and low temperatures. Oxygen, hydrogen, nitrogen, air, and 
carbon dioxide are examples of what are called perfect gases. 
They are fluids which require a very great reduction in tempera- 
ture and increase in pressure to bring them to the Hquid state. 
Vapors, on the other hand, are fluids which are readily trans- 
formed into Hquids by a very moderate reduction in tempera- 



.Connection fo itir Pump 




ture or increase in pressure. Com- 
mon examples of vapors with which 
engineers have to deal are steam and 
ammonia. 

For the present all our studies 

will be confined to the consideration 

of the properties of the perfect gases. 

Relation between Pressure and 

Fig. 2. — Constant Temperature Volume of a Perfect Gas. In prac- 

Apparatus for Demonstrating tically all heat engines, work is done 

Relation between Pressure and ^ ^j^ ^f ^^j^^^ ^f ^ ^^.^ ^^^ 

Volume of a Gas. ^ r 

the amount of work performed de- 
pends only on the relation of pressure to volume during such 
change and not at all on the form of the vessel containing this 
fluid. Since a good understanding of pressure and volume 
relations is most important, an illustration with a practical 
example will not be out of place. 

Fig. 2 shows a vessel filled with a perfect gas and surrounded 



lO ENGINEERING THERMODYNAMICS 

by a jacket filled with cracked ice. Its temperature will, 
therefore, be at 32° F. This vessel has a tightly fitting piston 
P of which the lower flat side has an area of one square foot. 
In the position shown the piston is two feet from the bottom of 
the vessel so that the volume between the piston and the bottom 
of the vessel is two cubic feet. The pressure on the gas is that 
due to the. piston and the weights shown. Assume this total 
weight is 100 pounds and that the air pump connected to the 
top of the vessel maintains a practically perfect vacuum above 
the piston. Then we say the pressure on the gas below the 
piston is 100 pounds per square foot. If now the weights are 
increased to make the pressure on the gas 200 pounds per 
square foot the piston will sink down until it is only one foot 
from the bottom of the vessel, provided the ice keeps a con- 
stant temperature.* 

Similarly, if the weight on the gas were 50 pounds and the 
vessel were made high enough, the lower side of the piston 
would be four feet from the bottom of the vessel. Examina- 
tion of these figures shows that for all cases the product of 
pressure and volume is constant, and in this particular case is 
always 200 foot-pounds. Similarly, if the volume is expressed by 
V in cubic feet per pound and the pressure by P in pounds per 
square foot the product is a constant quantity, which for air at 
32° F. is 26,220 foot-pounds. These facts are expressed by 
Boyle's Law, which states that the volume of a given mass of 
gas varies inversely with the pressure, provided the tempera- 
ture is kept constant; that is, V varies inversely as P or 

PF = a constant, (i) 

and also, if Pi and Vi represent some initial condition of pres- 
sure and volume and P2 and V2 are corresponding final con- 
ditions, then 

PiVi = P2F2 = a constant. (la) 

* If the temperature is not maintained constant, because of the tendency of 
gases to expand with increase in temperature, it will be necessary to apply a 
total weight greater than 200 pounds to reduce the volume to one cubic foot. 



PROPERTIES OF PERFECT GASES II 

Experiment shows that a permanent gas when heated in- 
creases in pressure if the volume is kept constant, and likewise 
increases in volume if the pressure is kept constant. If the 
pressure of a gas is kept constant and the temperature is raised 
1° F. the gas will expand 4^2 * ^f its volume at 32° F.; in 
other words, a volume of 492 cubic inches of gas will expand at. 
constant pressure to 493 cubic inches when the temperature is. 
raised from 32° to 33° F. 

Similarly, this same volume will be doubled if the gas is 
heated 492 degrees (i.e., to 524° F.) at a constant pressure; and 
on the other hand, if the volume remains constant and the 
pressure is allowed to vary, the same increase in temperature 
(492 degrees) will double the pressure which the gas had ini- 
tially. 

Conversely, if the same law could be held to apply at very 
low temperatures, at 492° below 32° F. or —460° F. ( — 273° C.)t 
the volume would be zero. 

Absolute Zero of Temperature. The examples above show 
that if temperatures are calculated, not from the ordinary zero 
but from 460 degrees below the ordinary Fahrenheit zero, the 
volume of the gas, if kept at constant pressure, will be propor- 
tional to the temperature reckoned from that zero point; and 
similarly, if the volume is kept constant, the pressure will be pro- 
portional to the temperature above that zero, which is called the 
absolute zero. Temperatures measured from the absolute zero 
as a basis are called absolute temperatures. 

Absolute Temperature Conversion. If we represent the tem- 
peratures on the ordinary Fahrenheit scale by t and the corre- 
sponding absolute temperatures by T then 

T = t-\- 460. (2) 

* This value is probably not exact. Various authorities give the following 

values: (i) — , (2)-^, (3)^—, (4) — , (5) ^-, (6) — ,etc. Thevalue 

492' 491-6' "" 491-4 493 492-7 49i 

given above is probably the best average value and is certainly close enough for 
all engineering calculations. 

t See end of chapter for a brief description of the latest low temperature re- 
search. 



12 ENGINEERING THERMODYNAMICS 

If /' is the temperature on the Centigrade scale and V the cor- 
responding absolute temperature then 

r = t' + 273. 

The laws of thermodynamics illustrated by the foregoing ex- 
amples, dealing with volume and pressure changes corresponding 
to temperature variations, may be stated in two short paragraphs 
as follows : 

(i) Under constant pressure the volume of a given mass of gas 
varies directly as the absolute temperature. 

(2) Under constant volume the absolute pressure of a given 
mass of gas varies directly as the absolute temperature. 

These fundamental principles, often called Charles' Laws, may 
also be stated thus : 

V T 
With pressure constant, — ^ = -7, (3) 

K 2 i 2 

P T 

With volume constant, — ^ = -7, (4) 

P2 T2 

where Vi and V2 are respectively the initial and final volimies, 

Pi and P2 are the initial and final absolute pressures, and Ti 

and T2 are the absolute temperatures corresponding to the 

pressures and volumes of the same subscripts. 

The following problem shows applications of Charles' laws: 

A gas has a volume of 2 cubic' feet, a pressure of 14.7 pounds 

per square inch absolute and a temperature of 60° F. 

(a) What will be the volume of this gas if the temperature is 
increased to 120° F., the pressure remaining constant? 

(b) What will be the pressure if the temperature is increased 
as in (a) but the volume remain constant? 

Solution, (a) Since pressure remains constant and the sub- 
stance is a gas, the volume varies directly as the absolute tem- 
perature. 

Letting Vi and Ti be the initial conditions and Vo and T2 be 
the final conditions, then 

Yl — Tl _^ _ 60 + 460 

V2 T2 Fo 120 + 460 

F2 = 2.33 cubic feet. 



PROPERTIES OF PERFECT GASES 13 

(b) Since the volume remains constant, 

?1 ^Xl 147 _ 60 + 460 ^ 

P2 T2 P2 120 + 460 

Pi = 16.39 pounds per square inch absolute. 

Absolute Pressure.* The ordinary types of pressure gages 
used in engineering work do not measure the pressure from a 
true basis for comparison. Such instruments measure only the 
excess of pressure above that of the atmosphere. The pressure 
indicated by the instrument is called gage pressure, and may 
be expressed as a '' true " or absolute pressure by adding the 
actual pressure of the atmosphere as obtained from the reading 
of a barometer. In all thermodynamic calculations in this 
book, unless it is expressly stated that gage pressures are meant, 
it is to be understood that absolute pressures are to be used. 
Boyle's and Charles' laws hold only for cases where the pressure 
P is in absolute terms. Furthermore, the same statement ap- 
plies to all rational formulas in thermodynamics because all 
have their fundamental basis on these laws. 

Gas Thermometers. If temperatures are measured by ther- 
mometers filled with a permanent gas like oxygen, hydrogen, air, 
or any other permanent gas used as the expansive medium, the 
increments of expansion are proportional to the temperature 

* It is frequently necessary to reduce the pressures in inches of mercury or of 
water to the equivalent in pounds per square inch. Since the weight of a cubic 
inch of mercury at 70° F. is 0.4906 pound and of water at the same temperature 
is 0.0360 pound, pressures in inches of mercury at the usual "room" temperatures 
can be reduced to pounds per square inch by multiplying by 0.491 or by dividing 
by 2.035; a^iid similarly, inches of water can be converted to pounds per square 
inch by multiplying by 0.0360 or by dividing by 27.78. Centimeters of mer- 
cury are reduced to pounds per square inch by multiplying by 0.1903. 

Kilograms per square centimeter are reduced to pounds per square inch by 
multiplying the kilograms per square centimeter by 14.223 or by dividing by 
0.0703. Grams divided by 28.35 s-^e ounces avoirdupois, or one gram is approxi- 
mately 3^0 ounce. 

A cubic foot of water at 70° F. weighs 62.3 pounds and at 32° F. 62.4 pounds. 
At the ordinary room temperature the pressure due to 2.31 feet of water is equiv- 
alent to one pound per square inch. 



14 ENGINEERING THERMODYNAMICS 

measured from the absolute zero. The air thermometer, some- 
times used by physicists, is one having the bulb and tube filled 
with air in such a way as to exert pressure upon a mercury col- 
umn in a manometer or " U-tube " attached to the end of the 
air tube. Expansions and contractions in volume of the air 
in the bulb will change the relative levels of the mercury in the 
manometer, which can be graduated to indicate temperatures. 

Combination of Boyle's and Charles* Laws. Equations (i), 
(3) and (4) cannot often be used in actual engineering problems 
as they stand, because it does not often happen that any one of 
the three variables {P, V and T) remains constant. A more gen- 
eral law must be developed, therefore, allowing for variations 
in all of the terms P, V and T. This is accomplished by com- 
bining equations (la), (3) and (4). It will be assumed that we 
are dealing with a pound of gas of which the initial conditions 
of pressure, volume and temperature are represented by Pi, Vi 
and Ti, while the corresponding final conditions, given by P2, 
V2 and T2, are arrived at in two steps. The first step is in chang- 
ing the volume from Vi to V2 and the pressure from Pi to 
some intervening pressure P2 while the temperature Ti remains 
constant. This change can be expressed by Boyle's law (equa- 
tion la). 

With constant temperature (Pi), 

(5) 

from which, by solving, P2' = " ^' ^ , (6) 

where P2 is the resulting pressure of the gas when its volume 
is changed from Vi to V2, with the temperature remaining 
constant at Pi. 

The second step is in the change in pressure from P2' to P2 
and the temperature from Pi to P2, while the volume remains 
constant at V2. This step is expressed by equation (4) as 
follows: 





P.' 


P/ 


PiFi 
F. ' 



PROPERTIES OF PERFECT GASES 



15 



With constant volume (F2), 

P2 Ti 



which may be written 



P2 T,' 
P2T2 



P, = 



Ti 



Substituting now the value of P2 from (6) in (8), we have 

which may be arranged to read, 

PiVi ^ P^ 



(7) 
(8) 

(9) 



(10) 



These steps are shown diagrammatically in Figs. 3 and 4. 
In Fig. 3 a surface is shown in which the lines indicating these 
changes lie. 




Fig. 3. — "Surface" Diagram Illustrating Derivation of "Combination" Law 

of Gases. 

The following problem shows the cippHcation of equation 10: 
A quantity of air at atmospheric pressure has a volume of 
2000 cubic feet when 'the barometer reads 28.80 inches of mer- 
cury and the temperature is 40° C. What will be the volume 



i6 



ENGINEERING THERMODYNAMICS 



of this air at a temperature of o° C. when the barometer reads 
29.96 inches of mercury? 




Letting 



Fig. 4. — Isometric Drawing Illustrating Derivation of "Combination" Law of 

Gases. 

Solution. Volume, pressure and temperature vary in this 
case as in the following equation, 

PiVi _ P2V2 

Pi, Fi, Ti = initial conditions, 
P2, F2, r2 = final conditions, 

28.80 X 2000 _ 29.96 X V2 
40 + 273 o + 273 ' 

F2 = 1676.42 cubic feet. 

Now, since P2, V2 and Ti in equation (10) are any other 
simultaneous conditions of the gas, we may also write the follow- 
ing more general relations: 

PiFi P2F2 PzVz 



then 



Ti 



and, therefore, 

where R is the " gas constant." 



T2 Ts 

PV = RT 



= a constant, 



(10') 
(11) 



PROPERTIES OF PERFECT GASES 1 7 

This constant is a most important quantity in thermody- 
namic calculations, its value varying only with the kind of gas 
dealt with. It must be remembered that equation (ii) was 
derived for one pound of gas, and therefore, in order to make 
it applicable for any mass of gas, it is only necessary to mul- 
tiply the constant R by the mass M, the volume V in the 
equation being the volume corresponding to the mass M. We 
thus obtain the final form of the equation representing the 
" combination law " of perfect gases: 

PV = MRT, (iiO 

where P = absolute pressure in pounds per square foot, 
V = actual volume in cubic feet, 
M = mass of gas in pounds, 

R = the "gas constant'' for one pound of gas in^ foot- 
pound units, 
T = the absolute temperature in Fahrenheit degrees. 

This equation is applicable to any perfect gas within the 
limits of pressure and temperature employed in common en- 
gineering practice. The ''thermodynamic" state of a gas is 
known when its pressure, volume, temperature, mass and com- 
position * are known, and when any four of these quantities 
are known the fifth can be found by equation (iiO- 

Values of the constant R in the table on page 23 have been 
calculated from experimental data of the specific volumes of the 
various gases mentioned. Thus, the specific volume of air is 
given as 12.38 cubic feet per pound at 32° F. (492 degrees 
absolute) and at atmospheric pressure (14.7 X 144= 21 17 pounds 
per square foot). Substituting these data in equation (iiQ, we 
have 

2117 X 12.38 = iXR X492, 
R = 53-3 (nearly). t 

It is also useful to observe that the value of R for any gas 

* The gas constant R depends upon the composition of the gas. 
t The tables on page 23 were calculated from slightly more exact data, but the 
agreement shown is good enough for all engineering calculations. 



l8 ENGINEERING THERMODYNAMICS 

can be approximately calculated by dividing the number 1544 
by its molecular weight. For example, the value of R for acet- 
ylene gas (C2H2), having a molecular weight of 24 + 2 or 26, is 
1544 -^ 26 or 59.4, which is the value commonly given for this 
gas. 

This latter method gives a very rapid and sufficiently exact 
means for determining the value of R of any gas for which the 
chemical formula is known. 

The following problem shows the application of equation 11': 

What is the volume of a tank that will hold a mass of 5 
pounds of air when the pressure is 200 pounds per square inch 
absolute and the temperature 40° C? 

Solution. Three unknowns are given and the fourth is to be 
found by the equation, 

PiVi = MiRiTi. 

Changing values given to the proper units, 
200 pounds per square inch absolute X 144 

= 28,800 pounds per square foot absolute. 

40° C. X I + 32 = 104° F. 
Then substituting in equation 11', 

28,800 X Fi = 5 X 53.3 (104 + 460); 
Vi = 5.22 cubic feet. 

Heat and Its Effect on Expansion. It is most important 
to discuss here, before going on to the more difficult prob- 
lems having to do with the expansions of gases, the most 
essential considerations involved in this study. In general the 
effect of adding heat to a gas is to raise its temperature. If 
heat is added so that at the same time the gas expands and 
does work its temperature may either rise or fall according as 
the amount of heat added is greater or less than the heat equiva- 
lent of the external work done. The following concise statement 
always appHes, and will be found very useful: 
Heat added = increase in internal energy -f- external work (12) 

Specific Heat. The amount of heat required to raise the 
temperature of a unit mass of a substance one degree is called its 



PROPERTIES OF PERFECT GASES 19 

specific heat. In the English system the specific heat is the 
number of British thermal units (B.t.u.) required to raise the 
temperature of a pound of the substance 1° F. 

The specific heat of gases and vapors changes considerably 
in value according to the conditions under which the heat is 
applied. If heat is applied to a gas or a vapor held in a closed 
vessel, with no chance for expansion, no work is performed, and, 
therefore, all the heat added is used to increase the temperature. 
This is the condition in a boiler, for example, when no steam is 
being drawn off. In this case we use the symbol C^ to represent 
the specific heat during its appHcation at constant volume. If, 
on the other hand, the heating is done while the pressure is kept 
constant and the volume is allowed to change to permit expan- 
sion and the performance of work,* we use the symbol Cp to 
represent the specific heat during its application at constant 
pressure. Heat application at constant pressure is the condition 
that is most interesting to the engineer. When his engines are 
running his boilers are making steam at constant pressure. 
The heat energy absorbed by a pound of a substance for raising 
only the temperature must certainly be the same regardless of 
the conditions of pressure and volume. Since, however, for 
constant pressure conditions some external work is always done, 
requiring a correspondingly larger amount of heat energy than 
for the case when the volume is constant, it follows that Cp is 
always greater than Cy. In other words, Cp is equal to Cy plus 
the heat equivalent of the work done by one pound of the sub- 
stance in expanding at constant pressure, while the tempera- 
ture is raised 1° F., if Cp and C^ are in B.t.u. 

Specific heat can also be expressed in units of work (foot- 
pounds). When Cp and Cv are expressed in B.t.u. then the 
corresponding values in foot-pounds will be 778 Cp and 
778 Q. 

External Work. The external work or the work done by a 
gas in its expansion is represented graphically by Fig. 5. This 

* Work performed as the result of expansion of a gas or vapor is always done 
at the expense of an equivalent amount of heat energy. 



20 



ENGINEERING THERMODYNAMICS 



W//////////////////y/////('/. 



gE= 



Pi — 



///<><'/<'/// . ,'<'/<'<'{<'//<'/ 



is the simplest sort of figure to show that the area under the 

expansion line BC is proportional to the work done in the expan- 
sion. Let us represent the initial 
condition of the gas at B as regards 
pressure and volume by Pi and Vi 
and the final condition at C by Pi 
and V2 (expansion being at con- 
stant pressure), then elementary 
mechanics teaches that the force 
P moved through a distance rep- 
resented by the difference of ab- 
scissas (F2 — Fi) is a measure of 
the work done. Obviously the area 
under the line BC divided by the 

horizontal length (F2 — Fi) is the average value of the force P. 

If, further, and in general, we represent the area under BC by 

the symbol A^ then we can write, 
A 



V: 



Fig. 5.- 



Volume 

External Work by Expan- 
sion. 



F2-F1 



= average value of P, 



whether or not P is constant. And also, 

A 



Work done = 



F2-F1 



X (F2 - V,) = A. 



The same principle applies whether the Kne BC is a straight 
line as shown or a very irregular curve, as will be shown later. 

Internal Energy, The heat energy possessed by a gas or vapor, 
or, we may say, the heat energy which is in a gas or vapor in 
a form similar to " potential " energy, is called its internal 
energy.* Thus, an amount of heat added to a substance when 
no work is performed is all added to the internal energy of that 
substance. On the other hand, when heat is added while work 
is being performed, the internal energy is increased only by the 
difference between the heat added and the work done. 

Internal energy may also be defined as the energy which a 

* It is also sometimes called intrinsic energy, since it may be said to "reside" 
within the substance and has not been transferred to any other substance. 



PROPERTIES OF PERFECT GASES 21 

gas or vapor possesses by virtue of its temperature, and may 
be expressed as follows: 

Internal energy = C^T (in B.t.u.), 

where T is the absolute temperature and Cv the specific heat 
at constant volume. From the paragraph on specific heat it 
will be remembered that Cv takes into account only that heat 
required to raise the temperature, since under constant volume 
conditions no external work is done; and therefore, in dealing 
with internal energy, since we do not care anything about the 
external work that may have been done, Cv is always used. 

It is with the change in internal energy that we are generally 
concerned, and this change is, obviously. 

Increase in internal energy 

= Cv {T2 — Ti) in B.t.u. (for one pound of gas). (13) 

This formula will always apply, for in whatever way the 
temperature of a substance is changed from Ti to T2 the change 
in its internal energy is the same since, by definition, it depends 
only upon the tempera ture^ 

Joule's Law. In the case of ideally perfect gases such as 
these thermodynamic equations must deal with, it is assumed, 
when a gas expands * without doing external work and with- 
out taking in or giving out heat (and, therefore, without 
changing its stock of internal energy) that its temperature does 
not change. 

Relation of Specific Heats and Gas Constant. In order to 
derive this relation we must consider certain fundamental equa- 
tions in thermodynamics and in general physics. 

A formula in physics for the total heat added, H, is as follows: 

H = mass X specific heat X difference in temperature. 

* It was for a long time supposed that when a gas expanded without doing 
work, and without taking in or giving out heat, that its temperature did not change. 
This fact was based on the famous experiments of Joule. Later investigations 
by Lord Kelvin and Linde have shown that this statement is not exactly correct 
as all known gases show a change in temperature under these conditions. This 
change in temperature is known as the "Joule-Thomson " effect. 



22 ENGINEERING THERMODYNAMICS 

If the heat is added at constant pressure, then obviously, in 
B.t.u., 

H = MC^{T,-T,). (14) 

Also, by equation (13), the increase in internal energy when 
heat is added = MC^{T2-Ti). (15) 

Another formula in physics is stated as follows: 

Work (external) = force X distance = Fd 

= PAd, where F= force in pounds, 
d = distance moved in feet, 
P = unit pressure in pounds per square foot, 
and A = area in square feet. 

Combining the factors A and d we obtain the change of 
volume, 

Ad = Av, where A^; is the change in volume. 

Therefore, External work = PAv. 

When the pressure remains constant, equation (15) may be 
expressed as follows: 

External work = P (V2 — Vi), foot-pounds 

Remember the fundamental equation (12), which is as follows: 
Heat added = increase in internal energy + external work. 
Substituting in the above equation (14), (15), and (16), we 
have 

MC, {r, - Td = MQ (T, - T,) + ^^\~^'^ ' (17) 

778 

By equation (iiO> 

PV2 = MRT2 and PVi = MRTi. 
Substituting these values ii^i (17), 

MC, (T, - ro = Mc. (r. - r,) + ^^^J^"^'^ • (is) 

778 

Simplifying, Cp = Cv + — r- (19) 

778 



PROPERTIES OF PERFECT GASES 



23 



Ratio of Specific Heats (7). The ratio of the specific heat at 
constant pressure to the specific heat at constant volume enters 
into many thermodynamic equations and is of so much im- 
portance in engineering calculations that it is commonly repre- 
sented by the symbol 7; thus 






(20) 



From the previous discussion of the relative values of Cp and C„ 
it is evident that 7 must always have a value greater than unity. 
By means of equations (19) and (20), the relation between 
Cv, 7 and R may be obtained: 

R 



C/p — Ov I 



From (b), 
Substituting in (a) 






Therefore, 



778 



Cv (7 - i) = 



a 



778 

R 



778 



R 



778 (7 - i) 



(a) 
(b) 



(21) 



DATA FOR VARIOUS GASES 



Name of gas. 



Air (pure) 

Carbon dioxide (CO2) .... 
Carbon monoxide (CO) . . . 

Hydrogen (H2) 

Marsh gas (CH4) 

Nitrogen (N2) 

Oxygen (O2) 

Sulphur dioxide (SO2) .... 



Molecu- 
lar 
weight. 



29 

44 
28 
2 
16 
28 
32 
64 



Spe- 
cific * 
volume, 
cu. ft, 
per lb. 



12.38 

8.17 
12.80 

179-65 
22.45 
12.81 
II .22 

5-61 



Den- 
sity,* 
lbs. per 
cu. ft. 



0.0807 
0.1224 
0.0779 
0.0056 
0.0445 
0.0779 
0.0891 
0.1781 



Value 
of R per 
lb. (ft.- 

Ib. 
units). 



S2> 
35 
55. 
766, 
96, 

55 
48, 

24 



0.237 
0.200 
o . 243 
3-4IO 

0-593 
0.244 
0.217 
0.154 



.169 
•154 
•173 
•435 
.470 
•173 
•155 
.123 



* Under standard conditions, Le., 32° F. and 14.7 lbs. per sq. in. absolute pressure. 



24 ENGINEERING THERMODYNAMICS 

LOW TEMPERATURE RESEARCHES 

In Harper^ s Monthly Magazine"^ is described the remarkable 
achievement of Professor Kamerlingh Onnes, of Leyden, in ap- 
proaching the absolute zero of temperature, viz., —273° C, 
within 1.8 degrees, when using helium. This is 93 degrees colder 
than the temperature of liquid air. Professor Onnes in 1913 
won the Nobel Prize for researches in the science of physics, 
and in 19 14 cooled a coil of lead wire to a temperature where 
the electrical resistance becomes zero, the current, once started, 
continuing to flow without supply so long as its temperature is 
kept at about 270 degrees below 0° C. 

At this low temperature many of the properties of matter at 
ordinary temperatures were expected to disappear, or at least 
to become greatly modified. The temperature of liquid air is 
— 180° C, and by its evaporation it is possible to reach —205° C. 
At these temperatures ordinary soft lead becomes hard and 
brittle like cast iron, rubber becomes fragile like glass, and 
alcohol can be frozen to a white solid and a candle can be 
made from it. The next lowest depression in temperature was 
made by Sir James Dewar, who, in liquefying hydrogen, attained 
a temperature of 253 degrees below 0° C, only 20 degrees from 
absolute zero of temperature, but no startHng effects were ob- 
served there. 

A reduction in temperature by evaporating a liquid with a 
very low boiling point results from the consumption of the 
heat within the liquid, making it necessary to prevent the 
entrance of heat from the outside. This is accomplished, firstly, 
by evaporating the liquid in a vacuum-jacketed glass vessel 
having the heat-insulating property of a '' thermos " bottle; sec- 
ondly, by surrounding the vessel with a Kquid of low temperature. 
Until recently the lowest temperature produced, about —260° C, 
was obtained by evaporating liquid hydrogen under reduced 
pressure in a vessel surrounded by liquid air. This tempera- 
ture was thought to be the lowest limit attainable, liquid hydro- 

* Vol. 129 (1914), pages 783-9. 



PROPERTIES OF PERFECT GASES 25 

gen having a lower boiling point than any other substance then 
known. 

However, when the peculiar and extremely rare gas known as 
helium was discovered, which is formed by the spontaneous de- 
composition of the element radium, the attainment of the ab- 
solute zero seemed feasible, provided a sufficient supply of 
helium could be obtained. This gas proved to be even more 
difficult to Hquefy than hydrogen, its boiling point being lower. 
However, it requires about 1000 cubic feet of gas to form one 
cubic foot of hquid. Professor Onnes succeeded in building an 
apparatus which enables him now to produce about half a pint 
of liquid heHum in a few hours. The glass vessel in which the 
helium is liquefied must be surrounded by a vessel containing 
liquid hydrogen, boihng under reduced pressure, which in turn 
must be surrounded by liquid air. The problem is comparable 
in difficulty to the construction of an apparatus for making 
ice inside of a furnace. Helium sells for about $50 a quart, 
but by the generosity of an American private commercial source, 
Professor Onnes obtained an enormous quantity after searching 
European markets in vain. 

Liquefaction of the gas helium is accomplished by means of 
an apparatus similar to that employed for the liquefaction of 
air, except that the compression pump, pipes and receiver must 
be absolutely gas-tight. Professor Onnes employs a pump with 
mercury pistons; every joint and valve in the system is im- 
mersed in a bath of oil, so that leaks will manifest themselves 
by small bubbles rising through the Hquid. The apparatus is 
sufficiently perfect to permit the compression of heliimi to 3000 
pounds per square inch. The compressed fluid is passed through 
pipes immersed in liquid air or hydrogen, then through the 
coiled tube of the Kquefier, whereby a portion becomes liquid 
and drops into a glass vessel, while the remainder passes off 
through another pipe and returns to the compressor — the proc- 
ess repeating itself over and over again without any loss of gas. 

At atmospheric pressi^re Hquid heHum boils at — 268.6° C. or 
4.4 degrees absolute; but by reducing the pressure in the tube 



26 ENGINEERING THERMODYNAMICS 

by means of an air pump, Professor Onnes has succeeded in 
reaching an actual measured temperature of only 1.8 degrees 
absolute. Such temperatures can be measured only by means 
of a gas thermometer filled with helium under very low pressure, 
instead of the usual hydrogen-filled thermometer. 

PROBLEMS 

1. The pressure in a gas tank is 90 lbs. per sq. in. by the gage. If 
atmospheric pressure is 15 lbs. per sq. in. absolute, what is the absolute 
pressure in the tank ? 

Ans. 105 lbs. per sq. in. 

2. What is the absolute pressure within the above gas tank in pounds 
per square foot? Ans. 15,120 lbs. per sq. ft. 

3. The gas in Problem i is at 60° F. temperature. What is the absolute 
Fahrenheit temperature of this gas ? Ans. 520 degrees. 

4. Air is at a temperature of 40° C. What is the absolute Fahrenheit 
temperature? Ans. 564° absolute F. 

5. Air at constant pressure has an initial volume of 2 cu. ft. and 
temperature of 60° F. ; it is heated until the volume is doubled. What is 
the resulting temperature in degrees Fahrenheit ? 

. J 1040° absolute F. 

JxnS . ■{ n n ^• t-« 

I 580 ordmary F. 

6. Air is cooled under constant volume. The initial condition is pres- 
sure of 30 lbs. per sq. in. absolute and temperature of 101° F. The final 
condition has a temperature of 50° F. What is the final pressure ? 

Ans. 27.27 lbs. per sq. in. absolute. 

7. One pound of hydrogen is cooled under constant pressure from a 
volume of i cu. ft. and temperature of 300° F. to a temperature of 60° F. 
What is the resulting volume? Ans. 0.684 cu. ft. 

8. A tank whose voliune is 50 cu. ft. contains air at 105 lbs. per sq. in. 
absolute pressure and temperature of 80° F. How many pounds of air does 
the tank contain ? Ans. 26.26 lbs. 

9. An automobile tire has a mean diameter of 34 in. and 4 in. width. 
It is pumped to 80 lbs. per sq. in. gage pressure at a temperature of 60° F.; 
atmospheric pressure 15 lbs. per sq. in. absolute. 

(a) How many pounds of air does the tire contain ? 

Ans. 0.38 lb. 

(b) Assuming no change of volume, what would be the gage pres- 

sure of the tire if placed in the sun at 100° F. ? 

Ans. 87 lbs. per sq. in. 



PROPERTIES OF PERFECT GASES 27 

10. An acetylene gas tank is to be made to hold 0.25 lb. of this substance 
when the pressure is 250 lbs. per sq. in. gage, atmospheric pressure 15 lbs. 
per sq. in. absolute, and the temperature of the gas 70° F. What will be 
the volume in cu. ft.? Ans, 0.206 cu. ft. 

11. A quantity of air at a temperature of 70° F. and a pressure of 15 lbs. 
per sq. in. absolute has a volume of 5 cu. ft. What is the volimie of the 
same air when the pressure is changed at constant temperature to 60 lbs. 
per sq. in. absolute? Ans. 1.25 cu. ft. 

12. How many pounds of air are present in Problem 11? 

Ans. 0.383 lb. 

13. The volume of a quantity of air is 10 cu. ft. at a temperature of 
60° F. when the pressure is 15 lbs. per sq. in. absolute. What is the pres- 
sure of this air when the volume becomes 60 cu. ft. and the temperature 
60° F.? Ans. 2.5 lbs. per sq. in. absolute. 

14. How many pounds of air are present in Problem 13 ? 

Ans. 0.779 lb. 

15. A tank contains 200 cu. ft. of air at a temperature of 60° F. and 
under a pressure of 200 lbs. per sq. in. absolute. 

(a) What weight of air is present? Ans. 207.5 lbs. 

(b) How many cubic feet will this air occupy at 15 lbs. per sq. 

in. absolute and temperature of 100° F. ? 

Ans. 2880 cu. ft. 

16. The volume of a quantity of air at 70° F. under a pressure of 15 lbs. 
per sq. in. absolute is 20 cu. ft. What is the temperature of this air when 
the volume becomes 5 cu. ft. and the pressure 80 lbs. per sq. in. absolute? 

Ans. 707° F. absolute. 

17. Air at constant pressure has a specific heat of 0.237 B.t.u. How 
many B.t.u. are required to raise 2 lbs. from 60° to 100° F. ? 

Ans. 18.96 B.t.u. 

18. Three pounds of a substance has 75 B.t.u. suppHed to it to change 
the temperature 100° F. What is its specific heat? Ans. 0.25 B.t.u. 

19. If the specific heat of air under constant pressure is 0.237 and the 
value of R is 53.3, find the value of the specific heat under constant volume. 

Ans. 0.169 B.t.u. 

20. From the data in Problem 19 find the value of 7 for air. 

Ans. 1.41. 

21. An auto tire has a volume of 0.66 cu. ft. and is to be pumped to a 
pressure of 85 lbs. per sq. in. gage at a temperature of 50° F. What vol- 
ume tank will be required to inflate four such tires if the air can be stored 
in this tank at 250 lbs. per sq. in. gage pressure and temperature of 70° F.? 

Suggestion. Assume the tires are filled with air at atmospheric pressure 
before starting to pump, and also when all tires have been inflated that 
the tank will contain its volume of air at the final pressure in the tires. 



28 ENGINEERING THERMODYNAMICS 

Let Pa = atmospheric pressure; Pt = gage pressure in tires; Pk = gage 
pressure in tank;* T = absolute temperature in tires or tank; Vt = vol- 
ume of tires ; Vk = volume of tank ; Ma = mass (pounds) of air in tires 
when at atmospheric pressure; Mb = total mass (pounds) of air in tires 
when at the tire pressure; Mc = total mass (poimds) of air in tank at final 
tire pressure; Md = total mass (pounds) of air in tank at the final tank 
pressure before beginning to fill the tires. 

Then 

(Mb — Ma) = Mass suppUed tires = Md — Mc 
iPt-hPa)Vt = MbXRxTt 

PaXVt = MaXRXTt 

PtXVt= (Mb - Ma) RTt 
Solving this for (Mb — Ma) : 

iPjc + Pa)Vk = MdXRXTk 
(Pt + Pa)Vk-McXRxTk 
(Pk-Pt)Vk=(Md-Mo)RTu. 

Substitute the value of (Mb — M,^ found above for (Md — Mc) and then 
solve for Vk- Ans. Vk = i-4i cu. ft. 

22. How many B.t.u. are required to double the volume of i lb. of air 
at constant pressure from 50° F.; specific heat is 0.237 B.t.u.? 

Ans, 121 B.t.u. 

23. A tank filled with 200 cu. ft. of air at 15 lbs. per sq. in. absolute and 
60" F. is heated to 150° F. 

(a) What will be the resulting air pressure in the tank ? 

Ans. 17.6 lbs. per sq. in. absolute. 

(b) How many B.t.u. will be required to heat the air? 

Ans. 240 B.t.u. 

24. A tank contains 200 cu. ft. of air at 60° F. and 40 lbs. per sq. in. 
absolute. If 500 B.t.u. of heat are added to it, what will be the resulting 
pressure and temperature? . 145-5 lbs. per sq. in. absolute. 

"''"^' 1 131.3° F. 

* All pressures are in pounds per square foot. 



CHAPTER III 
EXPANSION AND COMPRESSION OF GASES 

The general equation in the form PV = MRT"^ for the 
expansion or compression of gases has three related variables, 
(i) pressure, (2) volume and (3) temperature. For a given 
mass of gas with any two of these variables given, obviously, the 
third is fixed. As regards the analysis of the action of heat 
engines, the pressure and volume relations are most important, 
and graphical diagrams, called pressure- volume or P-V dia- 
grams, are frequently needed to assist in the analysis. The 



©100 

ft 



PV3 


PVi 


S^^ 


V 
I 1 


j 

1 ! 1 


— (p 

1 L.. 1 



P^ 2 4 6 8 10 

Volume, eu. ft. 

Fig. 6. — Diagram of Expansion and Compression at Constant Pressure. 

simplest sort of diagram of this kind is shown in Fig. 6, in 
which the vertical scale of coordinates represents pressures and 
the horizontal, volumes. 

Assume that the pressure and volume of a pound of a given 
gas are given by the coordinates P and Fi, which are plotted in 
the middle of the diagram. It will be assumed further that the 
pressure remains constant in the changes to be indicated. Now 
if the gas is expanded until its volume becomes V2, then its con- 
dition as regards pressure and volume would be represented at 
PV2. If, on the other hand, the gas had been compressed while 
a constant pressure was maintained, its final condition would 

* Equation (iiO, P- i7- 
29 



30 ENGINEERING THERMODYNAMICS 

be represented by the point PF3 to the left of PFi. Similarly, 
any line whether straight or curved extending from the initial 
condition of the gas at PVi will represent an expansion when 
drawn in the direction away from the zero of volumes and will 
represent a compression when tending toward the same zero. 

It is readily shown that areas on such diagrams represent the 
product of pressure and volume, and, therefore, work or energy 
(see page 20). Thus in Fig. 6 the area under the curve PF3 to 
PV2 represents on the scales given 100 (pounds per square foot) 
X (9 — i) cubic feet or 800 foot-pounds irrespective of whether 
it is an expansion or a compression from the initial condition. 
When a series of curves are joined together to form a closed 
figure which shows the varying conditions of pressure and vol- 
ume of a gas, the figure is called an indicator diagram. 

Most of the lines to be studied in heat engine diagrams are 
either straight or else they can be exactly or approximately 
represented by an equation in the form 

PF" = a constant, (22) 

where the index n, as experimentally determined, has varying 
numerical values, but is almost invariably constant for any 
one curve. When the lines of the diagram are straight the 
areas of simple rectangles and triangles need only be calculated 
to find the work done, and for these cases no discussion is here 
necessary. The two most common forms of curves to be dealt 
with in expansions are (i) when there is expansion with addi- 
tion of heat at such a rate as to maintain the temperature of 
the gas constant throughout the expansion. Such an expansion 
is called isothermal. The other important kind of expansion 
(2) occurs when work is done by the gas without the addition 
or abstraction of heat. To do this work some of the internal 
heat energy contained in the gas must be transformed in propor- 
tion to the amount of work done. Such an expansion is called 
adiabatic. 

The following problems show the application of the foregoing 
theory to straight line expansion: 



EXPANSION AND COMPRESSION OF GASES 31 

Case I. One pound of air having an initial temperature of 
60° F. is expanded to 100° F. under constant pressure. Find 

(a) External work during expansion; 

(b) Heat required to produce the expansion. 

Solution. From the fundamental theory, the heat added 
equals the increase in internal energy plus the external work 
done. In solving then for the heat added or required during 
any expansion it is only necessary to find the external work 
(which is equal to the area under the expansion curve) and add to 
it the heat needed to increase the internal energy. 

The external work W = Pi (V2 — Vi) or its equivalent 
MR{T2 -Ti)= 1 X 53.3 (100 + 460) - (60 + 460) = 2132 ft.-lbs. 

The increase in internal energy 

= Ma {T2 - Ti) 

= I X 0.169 (100 + 460) — (60 + 460) 
= 6.75 B.t.u. 

21^2 
Heat required = 6.75 + — ^ = 9.50 B.t.u. 

778 

As part of the data Cp is known. Then for this case of con- 
stant pressure expansion, the heat required equals by another 
method, 

MCp {T2 - Ti) = I X 0.237 (100 + 460) - (60 + 460) 
= 9.5 B.t.u. approximately. 
Case 2. One pound of air having an initial temperature of 
60° F. is heated at constant volume until the final temperature 
is 120° F. Find 

(a) External work; 

(b) Heat required. 

Solution. Applying the same theory as above, 
Heat added = increase in internal energy + external work. 
External work = o, since there is no area under the curve. 
Then 

Heat added = increase in infernal energy + o 
= MC. (r2 - Ti) + o 
= I X o'i69 (100 + 460) — (60 + 460) + o 
= 6.76 B.t.u. 



32 



ENGINEERING THERMODYNAMICS 



Isothermal Expansion and Compression. In an isothermal 
expansion or compression the temperature of the working sub- 
stance is kept constant throughout the process. When the 
temperature of the gas is kept constant, while the pressure and 
volume change, Boyle's Law (page lo) applies and we have 
simply 

PV = C = a constant. (23) 

This is the equation of a curve which is known in analytic 
geometry as a rectangular hyperbola. It is the special case of 
the general equation FV"" = constant (22), in which the index 
n = 1. If various values are substituted for V in equation (23) 
and the values found together with the corresponding calculated 




Volume 
Fig. 7. — -Work done by Isothermal Expansion and Compression. 

values of P are plotted, a curve Hke the one in Fig. 7 is obtained. 
An equation for the work done in the isothermal expansion of a 
gas will now be determined, starting with the initial condition 
of pressure and volume represented by Pi and Vi. The external 
work performed is shown graphically by the shaded area under 
the curve between A and B. Two vertical lines close together 
in the figure are the limits of a narrow closely shaded area and 
indicate an infinitesimal volume change dV, so small that the 
pressure may be assumed constant for the interval. Work 
done during this small change of volume is, then, 

dW = P dV, 



EXPANSION AND COMPRESSION OF GASES 33 

and for a finite change of volume of any size as from Vi to V2 
the work done, W (foot-pounds), is 



= J PdV. (24) 



w 

For integration of this form it is necessary to substitute P 
in terms of V. Assume that P and V are values of pressure 
and volume for any point on the curve of expansion of a gas of 
which the equation is 

PV = C (see equation 23). 
Then ^ " F* 

Substituting this value of P in equation (24), we have 

TF = C(logeF2-log. Fi). (25) 

Since the initial conditions of the gas are Pi and Vi, we have 
PV = C = PiVi, 
and substituting this value of C in equation (25), we obtain 

W = PiVi (loge V2 - loge Fi) 

or IF = Pi Fi loge— ^ (in foot-pounds). (26) 

Vi 

The above equation is quite general in application and can be 
used for any values of volume represented by Vi and F2. 

Units of mass do not enter. For the work done by a certain 
mass of gas under the same conditions we could write, since 

■ PiVi = MRT (in foot-pounds) and ]^ = §, 

Vi P2 

W = MRT loge ^ = MRT log, ^ (in foot-pounds) . (27) 
yi P2 

V2 / 
Often the ratio -f is called the ratio of expansion and is rep- 



34 ENGINEERING THERMODYNAMICS 

resented by r. Making this substitution we have, in foot- 
pounds, 

W = MRT loge r. (28) 

These equations refer to an expansion from PiFi to FiVi. 
If on the other hand, we wanted the work done in a compression 
from P\V\ to P3F3 the curve of compression would be from A 
to C and the area under it would be its graphical representa- 
tion. Equations (26), (27) and (28) would represent the work 
done the same as for expansion except that the expression 
would have a negative value; that is, work is to be done upon 
the gas to decrease its volume. 

The isothermal expansion or compression of a " perfect " gas 
causes no change in its stock of internal energy since T is constant 
(see page 20). During such an expansion the gas must take in 
an amount of heat just equal to the work it does, and conversely 
during such a compression it must reject an amount of heat 
just equal to the work spent upon it. This quantity of heat R 
(in B.t.u.) is, from equation (27), 

H = — — - loge — (m B.t.u.). (29) 

The following problem shows the application of the foregoing 
theory to isothermal expansions: 

Air having a pressure of 100 pounds per square inch absolute 
and a volume of i cubic foot expands isothermally to a volume 
of 4 cubic feet. Find 

(a) External work of the expansion; 

(b) Heat required to produce the expansion; 

(c) Pressure at end of expansion. 
Solution, (a) Since expansion is isothermal. 

External work, W = PiVi loge* ^ 

Vi 

= 100 X 144 X I X 1.3848 

= 19,941 foot-pounds. 

* 2.3 X log base 10 = log base e. Tables of natural logarithms are given on 
pages 195-196. 



EXPANSION AND COMPRESSION OF GASES 35 

(b) Since the heat added equals increase in internal energy 
plus external work, and since the temperature remains constant 
(requiring therefore no heat to increase the internal energy), the 
internal energy equals zero and the heat added equals the work 
done. 

Then Heat added = external work = ^'^'7 = 2t;.6 B.t.u. 

778 

(c) Since PiFi = P2F2, 
then 100 X I = ^2 X 4, 

P2 = 25 pounds per square inch absolute. 

If a gas expands and does external work without receiving a 
supply of heat from an external source, it must derive the amount 
of heat needed to do the work from its own stock of internal 
energy. This process is then necessarily accompanied by a low- 
ering of temperature and the expansion obviously is not iso- 
thermal. 

Adiabatic Expansion and Compression. Another most im- 
portant mode of expansion and compression from the viewpoint 
of the engineer is that in which the working substance neither 
receives nor rejects heat as it expands or is compressed and is 
called adiabatic. A curve which shows the relation of pressures 
to volumes in such a process is called an adiabatic line (see 
Fig. 8). In any adiabatic process the substance is neither 
gaining nor losing heat by conduction or radiation or internal 
chemical action. Hence the work which a gas does in such an 
expansion is all done at the expense of its stock of internal 
energy, and the work which is done upon a gas in such a com- 
pression all goes to increase its stock of internal energy. We 
could secure ideally adiabatic action if we had a gas expanding, 
or being compressed, in a cylinder which in all parts was a per- 
fect non-conductor of heat. The compression of gas in a cylin- 
der is approximately adiabatic when the process is very rapidly 
performed, but when done so slowly that the heat has time to 
be dissipated by conduction the compression is more nearly 
isothermal. Fig. 8 shows on a pressure- volume diagram the 



36 



ENGINEERING THERMODYNAMICS 



difference between an isothermal and an adiabatic for expansion 
or compression from an initial condition PiVi at A to final 
conditions at B and C for expansions, and at D and E for com- 
pressions. 




Volume 
Fig. 8. — Isothermal and Adiabatic Expansion Lines. 

Derivation of Equation for Adiabatic Expansion (or Com- 
pression). In order to derive the pressure- volume relation for 
a gas expanding adiabatically, consider the fundamental equa- 
tion (page 1 8), 

Heat added = increase in internal energy + external work or 

H = MK. {T2 -Ti)+P dV (foot-pounds), (30) 

where K^, is the specific heat in foot-pound units, i.e., 778 C^. 
Now in adiabatic expansion no heat is added or taken away 
from the gas by conduction or radiation, and, therefore, the 
left hand member of the above equation becomes zero. Fur- 
thermore, since the combination law (page 17) of Boyle and 
Charles can always be applied to perfect gases, the following 
simultaneous equations may be written: 

o = MK,dT + PdV, (31) 

PV = MRT. (32) 

When P, V and T vary, as they do in adiabatic expansion, 
equation (32) may be written as follows: 

PdV +VdP = MRdT, (33) 

and 

PdV-\-VdP 



dT = 



MR 



EXPANSION AND COMPRESSION OF GASES 37 

Substituting the value of dT in (31), we have 

^ ,,, , -,,^ PdV +VdP f . 

PdV + MK. -^ = o (34) 

RPdV + K,PdV + K.VdP ■■= o. 
To separate the variables divide hy PV: 

^dV ^ ^ dV ^ ^ dP , . 

Collecting terms, 

Integrating, 

{R + K,) log F + i:, log P = a constant = c. (36) 

^ + ^nog F + log P = c. 



R + K 



logPV ^^ =c. (37) 

Since from equation (19), 

R ~f" Kv = ivp, 
where Kp and Kv are respectively the specific heats at constant 
pressure and at constant volume in foot-pound units. 

Equation (37) becomes then 

Kp 

log PV^^ = c, 
and from equation (20), 

— ^ = 7 and, therefore, — ^ = 7, 

we may write 

PV = a constant. (38) 

Following the method used for obtaining an expression for 

the work done in isothermal expansion (equation 26), we can 

write again for the work done, W (in foot-pounds), for a change 

of volume from Vi to F2, 

■W^TpdV. (39) 



38 ENGINEERING THERMODYNAMICS 

Again we shall substitute, for purposes of integration, P in 
terms of V as outlined below. In the general expression PV"" 
= c, a, constant (see equation 22), where P and V are values of 
pressure and volume for any point on the curve of expansion 
of a gas of which the initial condition is given by the symbols 
Pi and Vi, we can then write, 

^=fn- (4°) 

And substituting (40) in (39), 

L—n + iJfi 

[ V^l-n _ 7 l-n -| 

Since PV"" = c = PiFi" = P2F2", we can substitute for c in 
(42) the values corresponding to the subscripts of V as follows: 

n — n 

l—n 

or W = ^"^^ ~ ^'^^ (foot-pounds). (43) 

n — I 

Since PV = MRT, 

W = ^^(^1-^^) (foot-pounds). (44) 

ft — I 

Equations (43) and (44) apply to any gas undergoing expan- 
sion or compression according to PV^ = sl constant. In the 
case of adiabatic expansion of a perfect gas n = y (see equation 

38). 

Change of Internal Energy During Adiabatic Processes. 

Since in adiabatic expansion no heat is conducted to or away 
from the gas, the work is done at the expense of the internal 
energy and, therefore, the latter decreases by an amount equiva- 



EXPANSION AND COMPRESSION OF GASES 39 

lent to the amount of work performed. This loss in internal 
energy is readily computed by equation (43) or (44). The 
result must be divided by 778 in order to be in B.t.u. 

During adiabatic compression the reverse occurs, i.e., there 
is a gain in internal energy and the same formulas apply, the 
result coming out negative, because work has been done on the 
gas. 

Relation between Volume, Pressure and Temperattxre in 
Adiabatic Expansion of a Perfect Gas. Since P, V and T vary 
during adiabatic expansion, it will be necessary to develop for- 
mulas for obtaining these various quantities. It will be remem- 
bered that equation (10), page 15, applies to perfect gases at 
all times. Therefore, in the case of adiabatic expansion or com- 
pression we can write the two simultaneous equations: 

^ = f^^- (A) 

By means of these two equations we can find the final condi- 
tions of pressure, volume and temperature, having given two 
initial conditions and one final condition. 

For instance, having given F1F2 and Ti, to find T2, divide (A) 
by (B)j member for member. Then 

Fi V2 



TiFi^ T2V2'' 



or 






In like manner the following formulas can be obtained: 



-© 



(46) 



40 ENGINEERING THERMODYNAMICS 

(47) 



-■(PJ 



(48) 

1 

^2 = Fi(g)', (49) 



= "'©'"■ 



(50) 



It should be noted that the above formulas can be used for 
any expansion of a perfect gas following PF" = a constant, 
provided 7 in the formulas is replaced by n. 

It is also to be noted that these equations can be used for 
any system of units so long as the same system of units is 
employed throughout an equation. 

The application of these equations to a practical example 

will now be shown by way of illustration. Take the case of a 

quantity of pure air in a cylinder at a temperature of 60° F. 

(Ti = 460 + 60 = 520 degrees absolute) which is suddenly (adi- 

Fi 2 
abatically) compressed to half its original volume. Then — - = -» 

V2 I 

and taking 7 from the table on page 23 as 1.40, the temper- 
ature immediately after compression is completed, T2 is calcu- 
lated by equation (45) as follows: 

T2 = Ti (pj = 520 i^-J = 520 X 20-40 = 688° absolute, 

or ^2 in ordinary Fahrenheit is 688 — 460 or 228 degrees. 

The work done in adiabatic compression of one pound of this 
air is calculated by equation (44) : 

^^. _ MR (T, - T,) _ 5.V3 (.^20 - 688) _ .=;.V3 ( - 168) _ 

7 — 1 1.40 — I 0.40 

foot-pounds per pound of air compressed. The negative sign 
means that work has been done on the gas. If the sign had 



EXPANSION AND COMPRESSION OF GASES 41 

been positive it would have indicated an expansion. As the 
result of this compression the internal energy of the gas has 

been increased by — '-"^ B.t.u., but if the cy Under is a con- 

778 

ductor of heat, as in practice it always is, the whole of this heat 
will become dissipated in time by conduction to surrounding 
air and other bodies, and the internal energy will gradually 
return to its original value as the temperature of the gas comes 
back to the initial temperature of 60° F. 

During the compression the pressure rises according to equa- 
tion (38): 

PV^ = constant, 

and just at the end the value is greater than the original pressure 
by the ratio of r^ to i,* or 2^-^^ to i, or 2.65 to i. If, as before, 
we now assume the temperature drops gradually to the initial 
temperature before compression (60° F.) without changing the 
volume, the pressure will fall with the temperature until it 
has at 60 degrees a value only twice as great as the original 
pressure. In other words, after cooling the pressure becomes 
inversely proportional to the change in volume produced by 
the compression. 

There are many cases of expansions which are neither adiabatic 
nor isothermal and which are not straight lines on P-V diagrams. 
It will, be observed from the equations in the discussion of the 
internal work done by an expanding gas and for the change of 
internal energy, that if in the general equation PF"* = a constant 
the exponent or index n is less than 7, the work done is greater 
than the loss in internal energy. In other words for such a 
case, the expansion lies between an adiabatic and isothermal and 
the. gas must be taking in heat as it expands. On the other 
hand, if n is greater than 7 the work done is less than the loss 
of internal energy. 

* Remember r is called the ratio of expansion and is f/- (See equation (28), 
page 34). 



42 ENGINEERING THERMODYNAMICS 

PROBLEMS 

1. How many foot-pounds of work are done by 2 lbs. of air in expand- 
ing to double its volume at a constant temperature of 100° F. ? 

Ans. 41,400 ft.-lbs. 

2. Three pounds of air are to be compressed from a volume of 2 to i cu. ft. 
at a constant temperature of 60° F. How many B.t.u. of heat must be 
rejected from the air? Ans. 74.2 B.t.u. 

3. An air compressor has a cylinder voliune of 2 cu. ft. If it takes air 
at 15 lbs. per sq. in. absolute and 70° F. and compresses it isothermally to 
100 lbs. per sq. in. absolute, find 

(a) Pounds of air in cylinder at beginning of compression stroke. 

Ans. 0.15 lb. 

(b) The final volume of the compressed air. Ans. 0.30 cu. ft. 

(c) The foot-pounds of work done upon the gas during compression. 

Ans. 8200 ft.-lbs. 

(d) The B.t.u. absorbed by the air in increasing the internal energy. 

Ans. o. 

(e) The B.t.u. to be abstracted from the cylinder. 

Ans. 10.46 B.t.u. 

4. Air at 100 lbs. per sq. in. absolute and a volume of 2 cu. ft. expands 
along ann = 1 curve to 25 lbs. per sq. in. absolute pressure. Find 

(a) Work done by the expansion. Ans. 39,900 ft.-lbs. 

(b) Heat to be supplied. Ans. 51.3 B.t.u. 

5. A quantity of air at 100 lbs. per sq. in. absolute pressure has a tem- 
perature of 80° F. It expands isothermally to a pressure of 25 lbs. per 
sq. in. absolute when it has a volume of 4 cu. ft. Find (i) the mass of air 
present, (2) Work of the expansion in foot-pounds, (3) Heat required in 
B.t.u. Ans. (i) 0.50 lb. 

(2) 19,950 ft.-lbs. 

(3) 25.6 B.t.u. 

6. Air at 100 lbs. per sq. in. absolute pressure and 2 cu. ft. expands to 
25 lbs. per sq. in. absolute adiabaticaUy. What is the final volume? 

Ans. 5.4 cu. ft. 

7. One cubic foot of air at 60° F. and a pressure of 15 lbs. per sq. in. 
absolute is compressed without loss or addition of heat to 100 lbs. per sq. 
in. absolute pressure. Find the final temperature and volume. 

Ans. 891° F. absolute; 0.257 cu. ft. 

8. Two pounds of air are expanded from a temperature of 300° F. to 
200° F. adiabaticaUy. How many foot-pounds of work are developed ? 

Ans. 26,300 ft.-lbs. 

9. A quantity of air having a volume of i cu. ft. at 60° F. under a pres- 
sure of 100 lbs. per sq. in. absolute is expanded to 5 cu. ft. adiabaticaUy. 



EXPANSION AND COMPRESSION OF GASES 43 

Find the pounds of air present, the final temperature of the air and how 
much work will be done during this expansion. Ans. 0.52 lb. 

273° F. absolute. 
16,900 ft. -lbs. 
10. Data same as Problem 3 but the compression is to be adiabatic. Find 

(a) The final volume of the compressed air. 

Ans. 0.516 cu. ft. 

(b) The final temperature of the compressed air. 

Ans. 911° F. absolute. 

(c) The foot-pounds of work to compress this air. 

Ans. 7510 ft.-lbs. 

(d) The B.t.u. absorbed by the air in increasing the internal 

energy. Ans. 9.65 B.t.u. 

(e) The B.t.u. to be abstracted from the gas. Ans. o B.t.u. 
II. A pound of air at 32° F. under atmospheric pressure is compressed 

to 4 atmospheres (absolute). What will be the final volume and the work 
of compression if the compression is (a) isothermal, (b) adiabatic ? 

Ans. (a) V2 = 3.09 cu. ft.; work = 33,800 ft.-lbs. 
(b) V2 = 4.60 cu. ft.; work = 31,500 ft.-lbs. 



CHAPTER IV 
CYCLES OF HEAT ENGINES 

When a gas or vapor undergoes a series of processes in which 
there is an interchange of heat quantities and is finally brought 
back to the condition, as regards its physical properties, which 
it had initially, the gas is said to have gone through or per- 
formed a cycle.* The most important cycle with which we have 
to deal is the Carnot cycle, because it is typical of the maximum 
efficiency obtainable. 

Carnot Cycle. Very important conclusions regarding theo- 
retically perfect heat engines are now to be drawn from the 
consideration of the action of an ideal engine in which the 
working substance is a perfect gas which is made to go through 
a cycle of changes involving both isothermal and adiabatic 
expansions and compressions. This ideal cycle of operations 
was invented and first explained in 1824 by Carnot, a French 
engineer, and gave us really the first theoretical basis for com- 
paring heat engines with an ideally perfect engine. For ex- 
plaining this Carnot cycle assume a piston and cylinder as 
shown in Fig. 9, composed of perfectly non-conducting material, 
except the cylinder-head (left-hand end of the cylinder) which is 
a good conductor of heat. The space in the cyhnder between 
the piston and the cylinder head is occupied by the working 
substance, which we shall assume to be a perfect gas. There 
is provided a hot body H of unlimited heat capacity, always 
kept at a temperature Ti, also a perfectly non-conducting cover 
N and a refrigerating or cold body R of unlimited heat receiving 
capacity, which is kept at a constant temperature T2 (lower 

* A thermodynamic machine performing a cycle in which heat is changed into 
work is called a heat engine, and one performing a cycle in which heat is trans- 
ferred from a medium at a low temperature to one at a higher temperature is 
called a refrigerating machine. 

44 



CYCLES OF HEAT ENGINES 



45 



than Ti). It is arranged that H, N or R can be apphed, as 
required, to the cyHnder head. Assume that there is a charge 
of one pound of gas in the cylinder between the piston and the 
cylinder head, which at the beginning of the cycle, with the piston 
in the position shown, is at the temperature T\, has a volume 
Va and has a pressure Pa> The subscripts attached to the 




Fig. 9. — Apparatus and Diagram Illustrating a Reversible (Camot) Cycle. 

letters V and P refer to points on the pressure- volume diagram 
shown in the figure. This diagram shows, by curves connecting 
the points a, b, c and d, the four steps in the cycle. 

The operation of this cycle will be described in four parts as 
follows: (i) Apply the hot body or heater H to the cylinder 
head * at the left-hand side of the figure. The addition of heat 

* It will be remembered that the head of this cylinder is a perfect conductor of 
heat. 



46 ENGINEERING THERMODYNAMICS 

to the gas will cause it to expand isothermally because the tem- 
perature will be maintained constant during the process at Ti. 
The pressure drops sHghtly to Ph when the volume becomes F?,. 
During this expansion external work has been done in advanc- 
ing the piston and the heat equivalent of this work has been 
obtained from the hot body H. 

(2) Take away the hot body H and at the same time attach 
to the cylinder head the non-conducting cover N. During this 
time the piston has continued to advance toward the right, 
doing work without receiving any heat from an external source 
so that the expansion of the gas in this step has been done at 
the expense of the stock of internal energy in the gas. The 
temperature has continued to drop * in proportion to the loss of 
heat to the value T^. Pressure is then Pc and the volume is Vc. 

(3) Take away the non-conductor N and apply the refriger- 
ator R. Then force the piston back into the cylinder. The 
gas will be compressed isothermally at the temperature T2. 
In this compression, work is being done on the gas, and heat is 
developed, but all of it goes into the refrigerator R, in which the 
temperature is always maintained constant at T^. This com- 
pression is continued up to a point d in the diagram, so selected 
that a further compression (adiabatic) in the next (fourth) 
stage will cause the volume, pressure and temperature to reach 
their initial values as at the beginning of the cycle.f 

(4) Take away the refrigerator R and apply the non-conduct- 
ing cover N. Then continue the compression of the gas without 
the addition of any heat. It will be adiabatic. The pressure 
and the temperature will rise and, if the point d has been prop- 
erly selected, when the pressure has been brought back to its 
initial value Pa the temperature will also have risen to its initial 
value Ti. The cycle is thus finished and the gas is ready for a 

* A pressure- volume diagram of a perfect gas does not show, graphically, changes 
of temperature. 

t Briefly the third stage of the cycle must be stopped when a point d is reached, 
so -located that an adiabatic curve {PV^ = constant) drawn from it will pass 
through the "initial" point a. 



CYCLES OF HEAT ENGINES 47 

repetition of the same series of processes comprising the 
cycle. 

To define the Carnot cycle completely we must determine 
how to locate algebraically the proper place to stop the third 
step (the location of d). During the second step (adiabatic 
expansion from b to c) we can write, by applying equation (45), 
the following temperature and volume relations: 



To 



=[f:r' 



also for the adiabatic compression in the fourth step we can 
state similarly, 

II 
T2 



Hence, 



[W'-m"- 



Simplifying and transposing, we have 

Observe that — is the ratio of expansion r (page 33) for the 

^ a 

isothermal expansion in the first step of the cycle. This has been 

shown to be equal to -f in the isothermal compression in the third 

yd 

step in order that the adiabatic compression occurring in the 
fourth step shall complete the cycle. 

A summary of the heat changes to and from the working 
gas (per pound) in the four steps of the Carnot cycle is as fol- 
lows: 

(ab). Heat taken in from hot body = RTi ^^ge— (by 

* a 

equation (2*8), in foot-pounds). (52) 

(be) . No heat taken in or rejected. 



48 ENGINEERING THERMODYNAMICS 

(cd). Heat rejected to refrigerator = i^r2 log* 7^ 

y c 

or — RT2 loge~ (by equation (28), in foot- 
ed 

pounds). (53) 

(da). No heat taken in or rejected. 

Hence, the net amount of work done, W, by the gas in this 
cycle, being the mechanical equivalent (foot-pounds), of the 
excess of heat taken in over that rejected, is the algebraic 'sum 

of (52) and (53): 

W = r(t, log. |-^ - T, log.|-j = R(T^- T,) loge^- (54) 

If the curves in Fig. 9 are accurately plotted to scales of pressure 
and volume, then the work in foot-pounds as calculated, from 
the measured area included in the P-V diagram will be found 
to agree exactly with the result given by equation (54) above. 

Efficiency of Carnot's Cycle. The thermal or heat efficiency 
of a cycle is usually defined as the ratio of 

Heat equivalent of work done 
Heat taken in 

The heat equivalent of work done is, by equation (54), 

V a 

and the heat taken in is, by equation (52), 

* a 

The ratio above representing the efficiency E is 



j?(rx-r.)iog.p 

* Log i^=- logy-- 



(ss) 



CYCLES OF HEAT ENGINES 49 

This efficiency represents the proportion of the total amount of 
heat given to the gas employed in a Carnot cycle, which the 
engine converts into work. At the temperature of the hot body 
H, the engine takes in an amount of heat proportional to the 
absolute temperature of this body or Ti, and in the course of 
the cycle rejects to the refrigerator R an amount of heat pro- 
portional to its absolute temperature 7*2 . The range of temper- 
ature for the cycle is, therefore, between Ti and T2. In the 
lowering of temperature corresponding to this range the engine 
converts into work that part of the heat taken in that is repre- 
sented by equation (55). Observe the conditions affecting the 
maximum efficiency of this cycle. For a given heat supply 
proportional to Ti the only way to increase efficiency is by 
reducing 7^2, so that the smaller T2 is the greater the tempera- 
ture range (T1 — T2) becomes and the higher the efficiency will be. 

Reversible Cycles. A heat engine which is capable of dis- 
charging to the "source of heat" when running in the reverse 
direction from that of its normal cycle the same quantity of 
heat that it would take from this source when it is running 
direct and doing work is said to operate with its cycle reversed, 
or, in other words, the engine is reversible. A reversible heat 
engine then is one which, if made to follow its indicator diagram 
in the reverse direction, will require the same horse power to drive 
it as a refrigerating machine as the engine will deliver when 
running direct, assuming that the quantity of heat used is the 
same in the two cases. An engine following Carnot's cycle is, 
for example, a reversible engine. The thermodynamic idea of 
reversibility in engines is of very great serviceableness because 
it will be shown that no heat engine can be more efficient than a 
reversible engine when both work between the same limits of 
temperature; that is, when both engines take in the same amount 
of heat at the same higher temperature and reject the same 
amount at the same lower temperature. 

Carnot's Principle. It was first proved conclusively by Car- 
not that no other heat engine can be more efficient than a re- 
versible engine when both work between the same temperature 



50 ENGINEERING THERMODYNAMICS 

limits. To illustrate this principle, assume that there are two 
engines A and B. Of these let us say A is reversible and B is 
not. In their operation both take heat from a hot body or 
heater H and reject heat to a refrigerator or cold body R. Let 
Qh be the quantity of heat which the reversible engine A takes 
in from the hot body H for each unit of work performed, and 
let Qr be the quantity of heat per unit of work which it dis- 
charges to the refrigerator R. 

For the purpose of this discussion, assume that the non- 
reversible engine B is more efficient than the reversible engine A. 
Under these circumstances it is obvious that the engine B will take 
in less heat than A and it will reject correspondingly less heat 
to R per unit of work performed. The heat taken in by the 
non-reversible engine B from the hot body H we shall designate 
then by a quantity less than & or Q^ — X and the heat rejected 
by B to the refrigerator R by Q^j. — X. Now if the non-reversible 
engine B is working direct (when converting heat into work) 
and is made to drive the reversible engine A according to its 
reverse cycle (when converting work into heat), then for every 
unit of work done by the engine B in driving the reversible 
engine A, the quantity of heat mentioned above, that is, Qu — X, 
would be taken from the hot body H by the non-reversible 
engine B and, similarly, the quantity of heat represented by Qu 
would be returned to the hot body H by the reverse action of 
the cycle of operations performed by A. This follows because 
the engine A is reversible and it returns, therefore, to H, when 
operating on the reverse cycle, the same amount of heat as it 
would take in from H when working on its direct cycle. By 
this arrangement the hot body H would be continually receiving 
heat, in the amount represented by X for each unit of work 
performed. At the same time the non-reversible engine B dis- 
charges to the refrigerator R a quantity of heat represented by 
Qji — X, while the reversible engine A removes from the refriger- 
ator R a quantity represented by Qr. As a result of this last 
operation the cold body will be losing continually per unit of work 
performed a quantity of heat equal to X. The combined per- 



CYCLES OF HEAT ENGINES 5 1 

formances of the two engines, one working direct as a normal heat 
engine and the other, according to its reverse cycle, as a com- 
pressor or what might be called a "heat pump," gives a constant 
removal of heat from the refrigerator R to the hot body H, 
and as a result a degree of infinite coldness must be finally pro- 
duced in the refrigerator. 

If we assume that there is no mechanical friction, this 
combined machine, consisting of a normal heat engine and 
compressor, will require no power from outside the system. 
For this reason the assumption that the non-reversible engine B 
can be more efficient than the reversible engine A has brought 
us to a result which is impossible from the standpoint of expe- 
rience as embodied in the statement of the " Second Law of 
Thermodynamics" (see page 3); that is, it is impossible to 
have a self-acting engine capable of transferring heat, infinite 
in quantity, from a cold body to a hot body. We should, 
therefore, conclude that no non-reversible engine, as B for 
example, can be more efficient than a reversible engine A when 
both engines operate between the same temperature limits. 
More briefly, when the source of heat and the cold receiver are 
the same for both a reversible heat engine and any other engine, 
then the reversible engine must have a higher possible efficiency; 
and if both engines are reversible it follows that neither can be 
more efficient than the other. 

Perfection in a Heat Engine. A reversible engine is perfect 
from the viewpoint of efficiency; that is, its efficiency is the best 
obtainable. No other engine than a reversible engine which 
takes in and discharges heat at identical temperatures will 
transform into work a greater part of the heat which it takes in. 
Finally, it should be stated as regards this efficiency that the 
nature of the substance being expanded or compressed has 
absolutely no relation to the thermal efficiency as outlined above. 

Reversed Carnot's Cycle. If an engine operating on Carnot's 
cycle is reversed in its action so that the same indicator diagram 
shown in Fig. 9 would Be traced in the opposite direction, the 
reversed cycle, when beginning as before at a with a per- 



52 ENGINEERING THERMODYNAMICS 

feet gas at the temperature Ti, will consist of the following 
stages : 

(i) When the non-conductor N is applied and the piston is 
advanced toward the right by the source of power performing 
the reversed cycle, the gas will expand, tracing the adiabatic 
curve ad, with constant lowering of temperature which at the 
point d will be T2. 

(2) When the non-conductor N is now removed, the refrig- 
erator R is applied, and the piston continues on its outward stroke. 
The gas will expand isothermally at the constant temperature Tij 
tracing the curve dc. During this stage the gas is taking heat 
from the refrigerator R. 

(3) When the refrigerator R is removed and the non-conductor 
N is again applied, which will be on the back stroke of the engine, 
the gas will be compressed, and on the indicator diagram another 
adiabatic curve cb will be traced. At the point b the temper- 
ature will be obviously Ti. 

(4) When the non-conductor N is removed and the hot body 
H is again applied, with the compression continuing along the 
isothermal curve ba, heat will be discharged to the hot body H, 
while the temperature is maintained constant at Ti. The cycle 
has now been traced in a reverse direction from the beginning 
back to the starting point at a, and is now complete. During 
this process no work has been done, but on the contrary an 
amount of work represented by the area of the indicator dia- 
gram, equivalent in foot-pounds to 

i^logeT^ (ri - To) (see equation (54), page 48), 

has been converted into heat. First, heat was taken from the 
refrigerator R, represented in amount by 

RT2\oge-zf^ 

y d 

and second, heat was rejected to the hot body H in the amount 
i^Tilog,^^ or -RTiloge — 

Vb Va 



CYCLES OF HEAT ENGINES 53 

As in direct operation of Carnot's cycle no heat is given or lost 

in the first and third stages outlined above. The algebraic sum 

Vb Vc 
of these two quantities, remembering that — = — , gives the 

'a ^ d 

net amount of work done, W, on the gas, and, therefore, the 
net amount of heat (foot-pound units) transferred from the cold 
body R to the hot body H or, 

W=RT,\oge~-RT,\oge^= -R\oge^{T,-T,). (56) 

'a 'a 'a 

Since the result is the same as given by equation (54), although 
opposite in sign on account of being work of compression, it 
will be observed that in the reverse cycle the same amount 
of heat is given to the hot body H as was taken from it in the 
direct operation of the same cycle, and that the same amount 
of heat is now taken from the refrigerator R as was in the other 
case given to it. 

Conclusions from the Above Discussion. In the explanation 
that has preceded of the performance of heat engines a perfect 
gas took in heat at the temperature of the source of supply of 
heat and discharged heat at the temperature of the refrigerator, 
or receiver of heat, the changes of temperature occurring as the 
result of adiabatic expansion or adiabatic compression. For a 
perfect gas we have determined, as in equation (55), that the 
efficiency of the heat engine was {Ti — T2) -^ T\. It has also 
been shown that Carnot's cycle was reversible, and that accord- 
ing to the "Second Law of Thermodynamics" no heat engine 
can have a higher efficiency than such a reversible engine when 
taking in and discharging heat at the same two temperatures 
Ti and T2. Finally, we are brought to the important conclusion 
that all reversible heat engines receiving and discharging heat 
at the same temperatures Ti and T2 are of equal efficiency, and 
that this efficiency, 

■£ = ^^^ (57) 



54 ENGINEERING THERMODYNAMICS 

having been determined for one reversible engine is also the 
efficiency of any other reversible engine, and is the maximum 
efficiency attainable with any engine. 

It should be observed in connection with the statement of an 
equation of efficiency like (57) above that it is impossible to 
utiKze the whole of any supply of heat for conversion into work 
because it is impossible to reach the absolute zero of tempera- 
ture, or in other words to make T2 in this equation practically 
zero. Considering only practical conditions, therefore, we may 
say that with given limits of temperatures Ti and T2 it is neces- 
sary for the attainment of the greatest efficiency that no heat 
shall be taken in by an engine except during the isothermal at 
the highest temperature and that no heat shall be rejected ex- 
cept at the isothermal at the lowest temperature T2. 

The following problem shows the application of the theory of 
Carnot's cycle: 

In Carnot's cycle the gas has an initial condition of 100 pounds 
per square inch absolute pressure, volume of i cubic foot and 
temperature of 300° F. The volume at the end of isothermal 
expansion is 2 cubic feet. Exhaust temperature is 60° F. Find 

(a) Heat supplied to the cycle; 

(b) Efficiency of the cycle; 

(c) Net work of the cycle. 
Solution. The heat supplied equals 

2 
100 X 144 X I X loge- = 9970 foot-pounds or 12.8 B.t.u. 

The efficiency is 

(^00 + 460) — (60 -f- 460) ^ ^ 4. 

^ —^ — ^ -^ — - = 0.316 or 31.6 per cent. 

300 -{- 460 

The net work of the cycle is 

9970 foot-pounds X 0-3^6 = 3150 foot-pounds. 

Regenerative Air Engines. The previous discussion has dealt 
entirely with ideal engines following what is known as Carnot's 
cycle. Such an engine has never been built. Another engine 
having theoretically a reversible cycle has, therefore, the same 



(CYCLES OF HEAT ENGINES 55 

practical application and will be next described. This engine, 
known as Stirling's, consists of two cylinders side by side, one 
of which is heated and the other cooled. It performs its cycle 
according to the following four stages: 

(i) Air which has been previously heated in a regenerator * 
to a temperature Ti is expanded isothermally from a volume 
represented by Vi to a volume F2. During this stage heat is 
being taken in from the furnace and the piston is raised as the 
expansion proceeds. The heat taken in during this stage in 
foot-pounds per pound of air is 

i^Ji loge— (see equation 52). 



V 



(2) During the next stage this air which is now highly heated 
is made to pass through the regenerator placed between the hot 
and the cold cylinders, and in its passage through the regenera- 
tor it gives up heat and has its temperature reduced to T2, with- 
out a change of volume. The heat absorbed by the regenerator 
is Cv (Ti — T2) . There is, of course, a drop in pressure corre- 
sponding to the reduction in temperature, although there is no 
change of volume. 

(3) In the cooled cylinder the air is compressed isothermally 
at the temperature T2 to its original volume. The heat dis- 
charged in this stage is 

RT2log^^= -RT2 log^- 

(4) The air is now passed back through the regenerator from 
the cooled cylinder to the heated cylinder, absorbing heat from 
the regenerator on the way and having its temperature raised 
to that at the beginning of the cycle or Ti. Heat taken in from 
the regenerator in this stage is Cv (Ti — T'^. 

* The regenerator consisting of a series of iron plates serves as a heat accumu- 
lator. When hot gases are passed through these plates they become heated. On 
the other hand, when cold gases are passed through the plates the heat previously- 
absorbed would be given up to the gas. 



56 



ENGINEERING THERMODYNAMICS 



The thermal efficiency of the whole cycle is then 
j^ _ Heat taken in — Heat discharged 
Heat taken in 
This would be expressed by the symbols already used as 



E = 



RTAogey - RT,\oge^ 



RTi loge 



Vi 



Canceling out the common terms, we have more simply 



E = 



Ti 



A theoretical indicator diagram from Stirling's engine is shown 
in Fig. 10. 




Volume 
Fig. io. — Indicator Diagram of Stirling's Air Engine. 

This engine, although not commercially a success, is, however, 
important because it represents the only type besides Carnot's 
that is reversible. The application of regenerators in heat 
engines is very limited and is confined almost exclusively to 
engines using hot air as the working substance. 

Another method of using air as the working substance in heat 
engines was developed by Ericsson, who used a cycle consisting 
of constant pressure and isothermal lines. The expansion took 
place at constant pressure while the air was passing through 
the regenerator. 

The following problem shows the application of the theory to 
cycles other than the Carnot. (Stirling engine.) 



CYCLES OF HEAT ENGINES 57 

Assume a cylinder of i cubic foot volume which contains air 
at 60° F. and 15 pounds per square inch absolute pressure and 
the cycle of operation to be performed as follows: 

(i) The gas is compressed adiabatically until the pressure 
equals 100 pounds per square inch absolute. 

(2) Heat is then supplied without change of volume of the gas 
and raises the temperature to 100° F. 

(3) The gas then expands adiabatically to a volume of i 
cubic foot. 

(4) Heat is then rejected without change of volume until the 
temperature is lowered to 60° F. The mass of air present is 

M= 15X144X1 3^^^ 

53.3 (60 + 460) 
The temperature at the end of adiabatic compression is 

7^2 = (60 + 460) j^^ = 895° F. absolute. 

The temperature after addition of the heat is 

Tz = 895 + 100 = 995° F. absolute. 

The volume after adiabatic compression is 

T/ 0.078 X 53.3 X 895 o -. 

V2 = — ^^-^ ^ = 0.258 cu. ft. 

100 X 144 

The temperature after adiabatic expansion equals 
T, = 995 i°-^^^y-'-' = 578° F. absolute. 

The heat supplied to the cycle is 

0.078 X 0.169 X 100 = 1.32 B.t.u. 

The heat exhausted from the cycle is 

0.078 X 0.169 X (578 — 520) = 0.76 B.t.u. 

The net work of the cycle is 

1.32 — 0.76 = 0.56 B.t.u. or 436 foot-pounds. 

The efficiency of the cycle is 

0-56 * 

— "^ = 0.42 or 42 per cent. 

1.32 



58 ENGINEERING THERMODYNAMICS 

PROBLEMS 

1. A Carnot engine containing lo lbs. of air has at the beginning of the 
expansion stroke a volume of lo cu. ft. and a pressure of 200 lbs. per sq. in. 
absolute. The exhaust temperature is 0° F. If 10 B.t.u. of heat is added 
to the cycle, find 

(a) Efficiency of the cycle. Ans. 15 per cent. 

(b) Work of the cycle. Ans. 1160 ft.-lbs. 

2. A Carnot cycle has at the beginning of the expansion stroke a pressure 
of 75 lbs. per sq. in. absolute, a volume of 2 cu. ft. and a temperature of 
200° F. The volume at the end of isothermal expansion is 4 cu. ft. The 
exhaust temperature is 30° F. Find 

(a) Heat added to cycle. Ans. 19.3 B.t.u. 

(b) Efficiency of cycle. Ans. 25.8 per cent. 

(c) Work of cycle. Ans. 3860 ft.-lbs. 

3. A cycle made up of two isothermal and two adiabatic curves has a 
pressure of 100 lbs. per sq. in. absolute and a volume of i cu. ft. at the 
beginning of isothermal expansion. At the end of adiabatic expansion the 
pressure is 10 lbs. per sq. in. absolute and the volume is 8 cu. ft. Find 

(a) Efficiency of cycle.* Ans. 20 per cent. 

(b) Heat added to cycle. Ans. 28.5 B.t.u. 

(c) Net work of cycle. Ans. 4360 ft.-lbs. 

4. In a steam power plant the steam is generated at 400° F. and is ex- 
hausted at 216° F. If the heat in the steam could be transformed accord- 
ing to Carnot's cycle, what would be the efiiciency of the plant ? 

Ans. 21.4 per cent. 

5. In a Carnot cycle the heat is added at a temperature of 400° F. and 
rejected at 70° F. The working substance is i lb. of air which has a volimie 
of 2 cu. ft. at the beginning and a volume of 4 cu. ft. at the end of isothermal 
expansion. Find 

(a) Volume at end of isothermal compression. Ans. 6.70 cu. ft. 

(b) Heat added to the cycle. Ans. 40.7 B.t.u. 

(c) Heat rejected from cycle. Ans. 25.2 B.t.u. 

(d) Net work of the cycle. Ans. 12,150 ft.-lbs. 

6. Air at a pressure of 100 lbs. per sq. in. absolute, having a volume of 
I cu. ft. and a temperature of 200° F., passes through the following opera- 
tions: 

15^. Heat is suppHed to the gas while expansion takes place under 
constant pressure untO the volume equals 2 cu. ft. 

2nd, It then expands adiabatically to 15 lbs. per sq. in. absolute 
pressure. 

* ^^., PiVi - P2V2 _ Ti-Ti • 



Efiiciency = 5-^7 or 



CYCLES OF HEAT ENGINES 59 

;^d. Heat is then rejected while compression takes place under con- 
stant pressure. 

4th. The gas is then compressed adiabatically to its original volume 
of I cu. ft. 
Find (a) Pounds of air used. Ans. 0.41 lb. 

(b) Temperature at end of constant pressure expansion. 

Ans. 1320° F. absolute. 

(c) Heat added to the cycle. Ans. 64.3 B.t.u. 

(d) Net work of cycle. Ans. 21,160 ft.-lbs. 

(e) Efficiency of cycle. Ans. 42.2 per cent. 



CHAPTER V 
PROPERTIES OF STEAM 

Steam in Heat Engines. The discussion on the preceding 
pages has had to do largely with the action of perfect gases in 
heat engines. Now we shall consider in this chapter a more 
limited field, confining ourselves exclusively to the action of 
water vapor or steam in such engines. The unusual physical 
properties of steam must, therefore, be explained here in con- 
siderable detail as well as the use and appHcation of tables of 
steam properties, called, for short, steam tables. Nearly all our 
dealings with steam in an engineering way will have to do with 
its formation at constant pressure. This is the condition of 
steam formation in a power plant boiler when the engines are 
at work. 

To make perfectly clear the process of steam formation in a 
boiler, assume that steam is to be made in a cylinder having a 
cross-sectional area of one square foot, closed at one end on 
which it stands. This cylinder is fitted with a frictionless piston 
(Fig. ii),* which is loaded so that it will exert a constant pres- 
sure of I GO pounds per square foot on the fluid in the cylinder 
below the piston. To begin the explanation of the properties 
of water vapor or steam, assume that there is in the bottom of 
the cylinder a quantity of water, say one pound, at just 32° F. 
(not ice) . If, now, heat is applied to the bottom of the cylinder, 
it will pass through the walls of the cylinder and will enter the 
water where it will produce the following changes in three 
stages: 

(i) As the water takes in heat its temperature rises until a 
certain temperature t is reached, at which steam begins to form. 
The value of t depends on the particular pressure which the 

* Compare with Fig. 2, page 9. 
60 



PROPERTIES OF STEAM 



6l 



piston and its load exerts. Until the temperature t is reached 
there is nothing but water below the piston. 

(2) After heating the water to the temperature t corresponding 
to the particular pressure exerted, and then adding more heat, 
there is no further rise in temperature, but water vapor (steam)' 
begins to form. With this formation of steam there is a rapid 
increase in volume, and the frictionless piston which is sup- 









Perfect 
Vacuum 


\ 


A 




I 

Connection 
to Air Pump 






Weight ^ 




m 
M 


WMM 






Water 



Fig. II. 



•Simple Apparatus to Illustrate Pressure and Volume Relations of 
Steam. 



posed to exert a constant pressure will be raised. The forma- 
tion of steam at constant temperature and constant pressure 
continues throughout this stage until all the water is converted 
into steam. All the steam which is formed during this stage is 
said to be saturated. 

(3) If, after all the water has been evaporated into steam, 
still more heat is added and taken in, the volume will be still 
further increased and there will also be an increase in tempera- 
ture. The steam in this last stage is then said to be super- 
heated. To make clear this distinction between saturated and 
superheated steam, it should be added that if at a given pres- 
sure steam exists at a temperature higher than the temperature 
(t) " of saturation," it is said to be superheated. The difference 
between saturated and superheated steam may also be expressed 
by saying that if water (at the temperature of the steam) be 



62 ENGINEERING THERMODYNAMICS 

mixed with the steam some of this water will be gradually evapo- 
rated if the steam is superheated, but not if the steam is satu- 
rated.* 

The properties of saturated steam differ very much from those 
X)i the perfect gases which we have been stud3dng, but when the 
steam is superheated to a high degree its properties approach 
very closely those of an ideally perfect gas. It is most impor- 
tant to remember that saturated steam at any particular pres- 
sure is always at the same temperature, while, on the other hand, 
superheated steam can have any temperature higher than that 
corresponding to saturated steam at the same pressure. 

Relation of Temperature, Pressure and Volixme in Saturated 
Steam. The important relations of temperature, pressure and 
volume were first determined in a remarkable series of experi- 
ments conducted by a French engineer named Regnault, and 
it has been on the basis of his data, first pubHshed in 1847, 
that even our most modern steam tables are based. Later ex- 
perimenters have found, however, that these data were some- 
what in error, especially for values near the dry saturated 
condition. These errors resulted because it was difficult in the 
original apparatus to obtain steam entirely free from moisture. 

The pressure of saturated steam increases very rapidly as the 
temperature increases in the upper limits of the temperature 
scale. It is very interesting to examine a table of the proper- 
ties of steam to observe how much more rapidly the pressure 
must be increased in the higher limits for a given range of tem- 
perature. 

It should be observed that in most tables the pressure is almost 
invariably given in terms of pounds per square inch, while in 
nearly all our thermodynamic calculations the pressure must 
be used in pounds per square foot. 

Heat in the Liquid (Water) (h) . The essentials of the process 
of making steam have been above described in a general way. 

* It is not unusual at all to find in practice hot water existing indefinitely in 
the presence of superheated steam. See Moyer's Steam Turbines, page 262, and 
JPower Plant Testing by the same author, page 316. 



PROPERTIES OF STEAM * 63 

The relation of this process to the amount of heat required will 
now be explained. If a pound of water which is initially at 
some temperature to is heated at a constant pressure P (pounds 
per square foot) to the boiling point corresponding to this pres- 
sure and then converted into steam, heat will first be absorbed 
in raising the temperature of the water from to to t, and then 
in producing vaporization. During the first stage, while the 
temperature is rising, the amount of heat taken in is approxi- 
mately (t— to) heat units, that is, British thermal units (B.t.u.), 
because the specific heat of water is approximately unity and 
practically constant. This number of B.t.u. multiplied by 778 
gives the equivalent number of foot-pounds of work. For the 
purpose of stating in steam tables the amount of heat required for 
this heating of water, the initial temperature to must be taken at 
some definite value; for convenience in numerical calculations 
and also because of long usage, the temperature 32° F. is invari- 
ably used as an arbitrary starting point for calculating the 
amount of heat " 'taken in." The symbol h (or sometimes q) is 
used to designate the heat required to raise one pound of water 
from 32° F. to the temperature at which it is vaporized into 
steam. In other words, " the heat of the liquid " (h) is the 
amount of heat in B.t.u. required to raise one pound of water 
from 32° F. to the boiling point. 

It is obvious, therefore, that we can write the value of the 
heat absorbed by water in being raised to the steaming tem- 
perature (h) in B.t.u., approximately, by the formula 

h = t - 32 (in B.t.u.). (58) 

More accurate values of h, taking into consideration the varia- 
tion in the specific heat of water, will be found in the usual 
steam tables. During this first stage, before any steaming has 
occurred, practically all the heat applied is used to increase the 
stock of internal energy. The amount of external work done 
by the expansion of water as a liquid is practically negligible. 

Latent Heat of Evaporation (L). In the second stage of the 
formation of steam as described, the water at the temperature t, 



64 ENGINEERING THERMODYNAMICS 

corresponding to the pressure, is changed into steam at that 
temperature. Although there is no rise in temperature, very 
much heat is nevertheless required to produce this evaporation 
or vaporization. The heat taken in during this stage is the 
latent heat of steam. In other words, the latent heat of steam 
may be defined as the amount of heat which is taken in by a 
pound of water while it is changed into steam at constant pres- 
sure, the water having been previously heated up to the tem- 
perature at which steam forms. The symbol L (also sometimes 
r) is used to designate this latent heat of steam. Its value 
varies with the particular pressure at which steaming occurs, 
being somewhat smaller in value at high pressures than at low. 

External Work of Evaporation (E). A part of the heat taken 
in during the '' steaming " process is spent in doing external 
work.* Only a small part of the heat taken in is represented 
by the external work done in making the steam in the boiler, 
and the remainder of the latent heat (L) goes to increase the 
internal energy of the steam. The amount of heat that goes 
into the performing of external work is obviously equal to P (the 
pressure in pounds per square foot) times the change of volume 
occurring when the water is changed into steam. 

Example. At the usual temperatures of the working fluid in 
steam engines the volume of a pound of water is about g^ of a 
cubic foot. The external work, W, done in making one pound of 
steam having finally a volume of V (cubic feet) at a constant pres- 
sure P (pounds per square foot) may be written in foot-pounds: 

W = External work = P(V- q\). (59) 

This last equation can be expressed in British thermal units 
(B.t.u.) by dividing by 778. It is apparent also from this equa- 

* The external work done in the formation of steam at constant pressure would 
be illustrated by the apparatus used for explaining the three stages of steam 
formation on page 61. It should be observed that while heat is being added 
merely to raise the temperature of the water there is practically no movement 
of the piston as there is scarcely any change in volume, but when steam is being 
made the volume increases rapidly and the piston will rise high above its initial 
position. 



PROPERTIES OF STEAM 65 

tion that the external work done in making steam is less at low 
pressure than at high,* because there is less resistance to over- 
come, or, in other words, P in equation (59) is less. The heat 
equivalent of the external work is, therefore, a smaller propor- 
tion of the heat added at low temperature than at high. 

Total Heat of Steam (H). The heat added during the process 
represented by the first and second stages in the formation of a 
pound of steam, as already described, is called the total heat of 
saturated steam or, for short, total heat of steam, and is repre- 
sented by the symbol H. Using the symbols already defined, 
we can write, per pound of steam, 

H = h + L(B.t.u.). (60) 

In other words, this total heat of steam is the amount of heat 
required to raise one pound of water from 32° F. to the tempera- 
ture of vaporization and to vaporize it at that temperature 
under a constant pressure. 

Remembering that h for water is approximately equal to the 
temperature less 32 degrees corresponding to the pressure at 
which the steam is formed (t) , we can also write approximately, 

H = (t - 32) + L. (61) 

To illustrate that equation (61) is approximately correct, take 
the case of steam being formed in a boiler at an absolute pressure 
of 115 pounds per square inch. 

From the steam tables we find that the temperature / of the 
steam at this pressure is 338° F., the latent heat of vaporiza- 
tion is 880 B.t.u. per pound and the total heat of the steam is 
1 189 B.t.u. per pound. To check these values with equation 
(59)5 we have, by substituting values of L and t, 

^ = (338 - 32) + 880 = 1 186 B.t.u. per pound. 

* Although at the lower pressure the volum«j of a given weight of steam is 
greater than at a higher pressure, the change of pressure is relatively so much greater 
in the process of steam formq,tion that the product of pressure and change of 
volume, P (F2 — Fi), which represents the external work done, is less for low 
pressure steam than for high. 



66 ENGINEERING THERMODYNAMICS 

When steam is condensed under constant pressure, obvi- 
ously the process which we have called the " second stage " is 
reversed and the amount of heat equal to the latent heat of 
evaporation (L) is given up during the change that occurs in the 
transformation from steam to water. 

Internal Energy of Evaporation and of Steam. It was ex- 
plained in a preceding paragraph that when steam is forming 
not all of the heat added goes into the internal or ** intrinsic " 
energy of the steam, but that a part of it was spent in per- 
forming external work. If, then, we represent the internal 
energy of evaporation by the symbol 1^, we can write, similarly 
to equation (59), in B.t.u. per pound of steam, 

I^ = ^_p(K^_M*. (62) 

778 

This equation represents the increase in internal energy which 
takes place in the changing of a pound of water at the tem- 
perature t into steam at the same temperature. In all the 
formulas dealing with steam that we have used we adopt the 
state of water at 32° F. as the arbitrary starting point from 
which the taking in of heat was calculated. This same arbitrary 
starting point is used also in expressing the amount of internal 
energy in the steam. This is the excess of the heat taken in 
over the external work done in the process. The total internal 
energy (I^) of a pound of saturated steam at a pressure P in 
pounds per square foot is equal to the total heat (H) less the 
heat equivalent of the external work done; thus, 

I^ = H-P^^^^^- (63) 

778 

Such reference is made here to the internal energy of steam 
because it is very useful in calculating the heat taken in and 
rejected by steam during any stage of its expansion or com- 

* It must be remembered that whenever there is a product of pressure and 
volume the result is in foot-pounds. To combine such a result with other terms 
in B.t.u. we must divide by 778. 



PROPERTIES OF STEAM 67 

pression. It is well to recall the following brief and simple 
statement (equation 12, page 18) : 

Heat taken in = increase of internal energy + external work 
done. 

When we are deaHng with a compression instead of an expan- 
sion then the last term above (external work) will be a negative 
value to indicate that work is done upon the steam instead of 
the steam doing work by expansion. 

The following problem shows the calculation of internal energy 
and external work: 

Example. A boiler is evaporating water into dry and satu- 
rated steam at a pressure of 300 pounds per square inch abso- 
lute. The feed water enters the boiler at a temperature of 145° F. 

The internal energy of evaporation per pound of steam is 

7^ _ 2, _ -P (^ - bV) _ o^j.^^ 300 X 144 (1.551 - fiV) 
778 ' 778 

= 811.3 -85.3 = 726.0 B.t.u., 

or taken directly from saturated tables equals 726.8 B.t.u. 

The total internal energy suppHed above 32° F. per pound of 
steam is 

Ih =H- ^ ^^ ~ ^^'^^ = 1204.1 - 85.3 = 1118.8 B.t.u., 
778 

or taken directly from saturated tables equals 1118.5 B.t.u. 

External work done above 32° F. per pound of steam as calcu- 
lated from steam tables is 

H — Ih = 1204. 1 — 1118.5 = 85.6 B.t.u. 

External work of evaporation per pound of steam as calcu- 
lated from the steam tables is 

L - Il = 811.3 - 726.8 = 84.5 B.t.u. 

The external work done in raising the temperature of a pound 
of water from 32° F. to the boiling point is 

85.6 - 84.5 = I.I B.t.u. 



68 ENGINEERING THERMODYNAMICS 

The external work done in raising the temperature of a pound 
of water from 32° F. to 145° F. is 

300 X 144 (0.0163 - 0.01602*) ^ ooj g 1-^ 

778 

The external work done in forming the steam from water at 
145° F. is, then, 

84.5 + 1. 10 — o.oi = 85.59 B.t.u. 
as compared with 85.3 B.t.u. as calculated from 

778 

It is to be noticed that the external work done during the addi- 
tion of the heat of the liquid is small as compared with other 
values, and for most engineering work it is customary to assume 
that no external work is done during the addition of heat to the 
water. With this assumption 

Iu = h + L- ^^^~J^^ =H-{L-Il). 
778 

Steam Formed at Constant Volume. When saturated steam 
is made in a boiler at constant volume, as, for example, when the 
piping connections from the boiler to the engines are closed, then 
no external work is done, and all the heat taken in is converted 
into and appears as internal energy Ih of the steam. This 
quantity is less than the total heat H of steam, representing its 
formation at constant pressure, by quantity P {V — -^-q) -^ 778, 
where P represents the absolute pressure at which the steam is 
formed in pounds per square foot and V is the volume of a 
pound of steam at this pressure in cubic feet. 

Wet Steam. In all expansions studied thus far, deaHng with 
saturated steam, it has been assumed that the steaming process 
was complete and that the water had been completely converted 
into steam. Now in actual engineering practice it is not at all 
unusual to have steam leaving boilers which is not perfectly 

* See Table 6 in Marks and Davis' Steam Tables and Diagrams for volumes of 
water. 



PROPERTIES OF STEAM 69 

and completely vaporized; in other words, the boilers are sup- 
plying to the engines a sort of mixture of steam and water. 
This mixture we call wet steam. It is steam which carries 
actually in suspension minute particles of water, which remain 
thus in suspension almost indefinitely. The temperature of this 
wet steam is always the same as that of completely saturated 
steam as given in the steam tables so long as any steam remains 
uncondensed. The ratio of the weight of moisture or water in 
a pound of wet steam to a pound of completely saturated steam 
is called the degree of wetness ; and when this ratio is expressed 
as a per cent, we have then what we call percentage of moisture or 
** per cent wet"; thus, if in a pound of wet steam there is 0.04 
pound of water in suspension or entrained, we speak of the steam 
as being four per cent wet. Another term, called the quality of 
steam, which is usually expressed by the symbol x, is also fre- 
quently used to represent the condition of wet steam. Quality 
of steam may be defined as the proportion of the amount of 
dry or completely evaporated steam in a pound of wet steam. 
To illustrate with the example above, if there is 0.04 pound 
of water in a pound of wet steam; the quality in this case 
would be I — 0.04 or 0.96. In this case we say then that the 
quaHty of this steam is ninety-six one-hundredths. 

With this understanding of the nature of wet steam it is obvi- 
ous that latent heat of a pound of wet steam is xL. Similarly, 
the total heat of a pound of wet steam is h -{- xL, and the 
volume of a pound of wet steam is xV -h g^ (i — x) or approxi- 
mately equal to xV, because the term gV (^ ~ ^) i^ negligibly 
small except in cases where the steam is so wet as to consist 
mostly of water. Similarly, the internal energy in a pound of 
wet steam is 

l^ = h + x'^L-^^^^^]^- (64) 

Superheated Steam. When the temperature of steam is higher 
than that corresponding to saturation as taken from the steam 
tables, and is, therefore; higher than the standard temperature 
corresponding to the pressure, the steam is said to be super- 



70 ENGINEERING THERMODYNAMICS 

heated. In this condition steam begins to depart and differ 
from its properties in the saturated condition, and when super- 
heated to a very high degree it begins to behave somewhat 
like a perfect gas. 

There are tables of the properties of superheated steam just 
as there are tables of saturated steam. Tables of superheated 
steam are very much larger and cover many more pages than 
those of saturated steam, for the reason that for every pressure 
there are innumerable values for temperature and also for vol- 
ume. When dealing with saturated steam there is always only 
one possible temperature and only one specific volume to be 
considered. With superheated steam, on the other hand, for a 
given pressure we may have any temperature above that of 
saturated steam, and corresponding to each temperature there 
will be, of course, definite values for specific volume and total 
heat. Like a perfect gas the specific volume or the cubic feet 
per pound increases with the increase in temperature. 

Total heat of superheated steam is obviously greater than 
the total heat of saturated steam which is not wet by the amount 
of heat that must be added to dry * saturated steam to produce 
the required degree of superheat. Thus, since the total heat 
of dry saturated steam is, as before, h -\- L, the total heat of 
superheated steam with D degrees of superheat is 

Ks = h + L-\-CpXD; 
or if we call the temperature of the superheated steam /gup and 
/sat is the temperature of saturated steam corresponding to the 
pressure, we can write, similarly, 

"Bis = h -\- L -\- Cp (/gup — /gat ) • . 

The formulas above for superheated steam are for the total 
heat of steam at constant pressure as shown by the use of Cp. 
In practically all engineering calculations it is only the condition 
of total heat at constant pressure that interests us. 

* In practice we speak of steam as being dry saturated when it is exactly sat- 
urated and has no moisture. It is the condition known simply as saturated steam 
as regards the properties given in the ordinary steam tables. The dry saturated 
condition is the boundary between wet steam and superheated steam. 



PROPERTIES OF STEAM 7 1 

It should be carefully observed that the total heat of super- 
heated steam is the amount of heat required to produce a pound 
of steam with the required degrees of superheat from water at 
32° F. 

To obtain the amount of internal energy of a pound of steam 
(superheated) corresponding to this total heat as stated above, 
the external work expended in the steaming process must be 
subtracted; that is, 

Ih= h + L - ^^^-\~^'^^ +C,feup - 4at) - ^^^^^% ^"^^ 
778 778 

= h ^L+C, feup -4at) - ^ ^^^^^ 7 '^ ^ = Hs - -^ (^-up - /o) . 

778 778 

The term ^ can be neglected in the equations above for prac- 
tically all engineering calculations as the maximum error from 
this is not likely to be more than one in one thousand or ^0 per 
cent.* The accuracy of our steam tables for values of latent and 
total heat is not established to any greater accuracy. Making 
this approximation, the equation above becomes 

lH = h + L + C, (4up - U - (^^^^) = H. - 1^<^ . 

V 778 / 778 

To make these matters clearer, examine, for example, the prop- 
erties of superheated steam as given in Marks and Davis' Steam 
Tables f for superheated steam at 165 pounds per square inch 
absolute pressure and 150° F. superheat. We read as follows: 

Press, lbs. per Deg. of Sup, 

sq. in. abs. F. 

165 / 516.0 

y 3-43 

Hs or hX 1277.6 

* The error in the value of total internal energy due to neglecting the term -^q 
in the exercise worked out on page 67 is 0.85 B.t.u. per pound (one per cent of 
the external work), or an error of 0.76 per cent in the final result. 

t Page 48, line 13 (First Edition). 

t In their tables Marks and Davis represent the total heat of superheated steam 
by h. In this book the symbol Hs is used for greater clearness and to be con- 
sistent with the symbols used in the preceding formulas. They use also v for 
specific volume in place of V as above. 



72 



ENGINEERING THERMODYNAMICS 



The values of h and L for saturated steam at this pressure are 
respectively 338.2 and 856.8. From the curves given, Fig. 12,* 
we find that the specific heat of superheated steam at constant 
pressure (Cp) is 0.552. From these data we could obtain, then, 
^a = ^-f L + Cp Xi5o= 338.2 +856.8+ 0.552 X 150= 1277.7 


















f ^ 


r 


\ ^ 






























PC 














, 




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\ 




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\> 






























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\ 


> 












































\ 




v^ 


\\ 


\, 


























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/ 


\ 




\ 


\ 


s 


s 


y 






1 
















.58 










/ 




N 


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V, 


s\ 


\ 


'v 


s 






























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\ 


\ 


V 


S 


\ 


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s 




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s 


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i 




N 




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\ 


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"V 


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4.4 


Lb 


s.Abi 












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V 




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s. 


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J 


g 


Jii 












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■22 


7.2 
).0- 
0.4 
2.2 
3.6 










1 






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■->. 


^ 






— • 


— 


-17 










/ 
















•*~. 


^ 


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_ 


-11 










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- 


~U 









_ 


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» 










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_ 

















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V 


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_ 


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=5 


c 


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" 


























































































































.42 

















































Fig. 12. 



200 250 300 350 400 450 500 550 600 050 700 750 
Temperature ° F 

■Mean Values of Cp Calculated by Integration from Knoblauch 
and Jakob's Data. 



(nearly). Observe that this value agrees with that given in 
the tables as indicated above. The specific volume (F) is cal- 
culated from the empirical formula derived from experimental 
results and is expressed as follows: 

V = [0.5962 T-p{.+ 0.0014 p) [ ''°''^'°°° - 0.0833)] ^, 

* The curves for values of Cp, as given in Fig. 12, are for average and not for 
instantaneous values such as are- given in Fig. 13 and also in Fig. 5, page 97, of 
Marks and Davis' Tables and Diagrams. Great caution must be observed in the 
use of curves of this kind. Those giving instantaneous values can only be used 
for the value of Cp in the formulas given for the total heat of superheated steam 
after the average value has been found by integrating the curve representing 
these values for a given pressure. 



PROPERTIES OF STEAM 



73 



where p is in pounds per square inch, V is in cubic feet per pound 
and r = / + 460 is the absolute temperature on the Fahrenheit 
scale.* 



0.90 
0.85 
0.80 

0.75 

0.70 

c 

I 
|o.65 

I 

f^O.60 
0.55 
'^ 






20 






• 






1 


H 












IG 


1 












14/ 


1 










12^ 


\l 










10/ \ 


WW 










c/\ 


\vl 


Pressu 
sq. cm. a 


res in Kg 
bsolute. 


m. per 


/ 


K^ 


^ 


^ 








A 




"~^ 




::r — -^ 




I_— — ^ 


^^^^^ 


e.45 


— 


-^;;r^ 


^^^^^ 















100 150 200 250 ^300 350 400 

Temporature C. 

Fig. 13. — Values of the "True " Specific Heat of Superheated Steam. 

Drying of Steam by Throttling or Wire-drawing. When steam 
expands by passing through a very small opening, as, for example, 
through a valve only partly open in a steam line, the pressure 



* The value of 7 or -^ of superheated steam used in ordinary engineering calcu- 
lations is 1.3. 



74 ENGINEERING THERMODYNAMICS 

is considerably reduced. Steam engineers usually call this 
effect throttling or wire-drawing. The result of expansion of 
this kind when the pressure is reduced and no work is done 
is that if the steam is initially wet it will be drier and if it is 
initially dry or superheated the degree of superheat ^dll be in- 
creased. The reason for this is that the total heat required to 
form a pound of dry saturated steam {H) is considerably less at 
low pressure than at high, but obviously the total quantity of 
heat in a pound of steam must be the same after wire-drawing 
as it was before, neglecting radiation. Now if steam is initially 
wet and the quahty is represented by Xi, then the total heat in 
the steam is represented by hi plus XiLi, in which hi and Li repre- 
sent the heat of the Kquid and the latent heat of the steam at 
the initial pressure. If, also, the quality, heat of liquid and the 
latent heat of the steam after wire-drawing are represented 
respectively hy oc^, h^ and L2, then hi + XiLi = h. -\- x^Li. It 
happens in many engineering calculations that all of the terms 
in the last equation are known except x^ and this can be solved 
as shown in the following equation: 

_ XiLi -\- hi — h. 

x^ — 



This drying action of steam in passing through a small open- 
ing or an orifice is very well illustrated by steam discharging 
from a small leak in a high pressure boiler into the atmosphere. 
It will be observed that no moisture is visible in the steam a 
few inches from the leak but farther off it becomes condensed 
by loss of heat, becomes clouded and plainly visible. An im- 
portant application is also to be found in the throttling calo- 
rimeter (page 75). 

Determination of the Moisture in Steam. Unless the steam 
used in the power plant is superheated it is said to be either dry 
or wet, depending on whether or not it contains water in sus- 
pension. The general types of steam calorimeters used to de- 
termine the amount of moisture in the steam may be classified 
under three heads: 



PROPERTIES OF STEAM 75 

1. Throttling or superheating calorimeters. 

2. Separating calorimeters. 

3. Condensing calorimeters. 

Throttling or Superheating Calorimeters. The type of steam 
calorimeter used most in engineering practice operates by passing 
a sample of the steam through a very small orifice, in which it 
is superheated by throttling. A very satisfactory calorimeter of 
this kind can be made of pipe fittings as illustrated in Fig. 14. 
It consists of an orifice O discharging into a chamber C, inta 
which a thermometer T is inserted, and a mercury manometer 
is usually attached to the cock V3 for observing the pressure in 
the calorimeter. 

It is most important that all parts of calorimeters of this 
type, as well as the connections leading to the main steam pipe, 
should be very thoroughly lagged by a covering of good insu- 
lating material. One of the best materials for this use is hair 
felt, and it is particularly well suited for covering the more or 
less temporary pipe fittings, valves and nipples through which 
steam is brought to the calorimeter. Very many throttling 
calorimeters have been declared useless by engineers and put 
into the scrap heap merely because the small pipes leading to 
the calorimeters were not properly lagged, so that there was too 
much radiation, producing, of course, condensation, so that the 
calorimeter did not get a true sample. It is obvious that if the 
entering steam contains too much moisture the drying action 
due to the throttling in the orifice may not be sufiicient to super- 
heat. It may be stated in general that unless there is about 
5° to 10° F. of superheat in the calorimeter, or, in other words, 
unless the temperature on the low pressure side of the orifice is 
at least about 5° to 10° F. higher than that corresponding to the 
pressure in the calorimeter, there may be some doubt as to the 
accuracy of results.* The working limits of throtthng calo- 

* The same general statement may be made as regards determinations of 
superheat in engine and turbin^ tests. Experience has shown that tests made 
with from o to 10 degrees Fahrenheit superheat are not reliable, and that the 
steam consumption in many cases is not consistent when compared with results 



76 ENGINEERING THERMODYNAMICS 

rimeters vary with the initial pressure of the steam. For 35 
pounds per square inch absolute pressure the calorimeter ceases 
to superheat when the percentage of moisture exceeds about 
2 per cent; for 150 pounds absolute pressure when the moisture 
exceeds about 5 per cent; and for 250 pounds absolute pressure 
when it is in excess of about 7 per cent. For any given pressure 
the exact limit varies slightly, however, with the pressure in the 
calorimeter. 

In connection with a report on the standardizing of engine 
tests, the American Society of Mechanical Engineers* published 
the following instructions regarding the method to be used for 
obtaining a fair sample of the steam from the main pipes. It is 
recommended in this report that the calorimeter shall be con- 
nected with as short intermediate piping as possible with a 
so-called calorimeter nipple made of |-inch pipe and long enough 
to extend into the steam pipe to within | inch of the opposite 
wall. The end of this nipple is to be plugged so that the steam 
must enter through not less than twenty J-inch holes drilled 
around and along its length. None of these holes shall be less 
than I inch from the inner side of the steam pipe. The sample 
of steam should always be taken from a vertical pipe as near as 
possible to the engine, turbine or boiler being tested. A good 
example of a calorimeter nipple is illustrated in Fig. 15. 

Never close and do not usually attempt to adjust the discharge 
valve V2 without first closing the gage cock V3. Unless this 
precaution is taken, the pressure may be suddenly increased in 
chamber C, so that if a manometer is used the mercury will be 
blown out of it; and if, on the other hand, a low-pressure steam 
gage is used it may be ruined by exposing it to a pressure much 
beyond its scale. 

Usually it is a safe rule to begin to take observations of tem- 
perature in calorimeters after the thermometer has indicated a 

obtained with wet or more highly superheated steam. The errors mentioned 
when they occur are probably due to the fact that in steam, indicating less than 
10 degrees Fahrenheit superheat, water in the hquid state may be taken up in 
*' slugs" and carried along without being entirely evaporated. 

* Proceedings American Society of Mechanical Engineers, vol. XXI. 



PROPERTIES OF STEAM 77 

maximum value and has again slightly receded from it. The 
quahty or relative dryness of wet steam is easily calculated by 
the following method and symbols : 

pi = steam pressure in main, pounds per square inch 

absolute, 
p2 = steam pressure in calorimeter, pounds per square 

inch absolute, 
tc = temperature in calorimeter, degrees Fahrenheit, 
Li and hi = heat of vaporization and heat of liquid corre- 
sponding to pressure ^1, B.t.u., 
^2 and /2 = total heat (B.t.u.) and temperature (degrees 
Fahrenheit) corresponding to pressure ^2, 
Cp = specific heat of superheated steam. Assume 0.47 

for low pressures existing in calorimeters, 
X\ = initial quality of steam. 

Total heat in a pound of wet steam flowing into orifice is 

XiLi + ^1, 

and after expansion, assuming all the moisture is evaporated, 
the total heat of the same weight of steam is 

H2 + Cp [tc — h). 

Then assuming no heat losses and putting for Cp its value 0.47 we 
have 

ociLi + hi = H2 + 0.47 {tc - k) 

X ^ H2 + 0.47 {ic -h) -hi .^ s 

Li 

The following example shows the calculations for finding the 
quality of steam from the observations taken with a throttling 
calorimeter : 

Example. Steam at a pressure of loc pounds per square inch 
absolute passes through a throttling calorimeter. In the calo- 
rimeter the temperature Of the steam becomes 243° F. and the 
pressure 15 pounds per square inch absolute. Find the quahty. 



78 ENGINEERING THERMODYNAMICS 

Solution. By taking values directly from tables of properties 
of superheated steam,* the total heat of the steam in the calorim- 
eter at 15 pounds per square inch absolute pressure and 243° F. 
is 1 164.8 B.t.u. per pound or can be calculated as follows: 

H + Cp {to- k) = 1150.7 + 0.47 (243 - 213) 
= 1 164.8 B.t.u. per pound. 

(Note, tc = temperature in calorimeter and ^2 = temperature 
corresponding to calorimeter pressure.) 

The total heat of the steam before entering the calorimeter 
is h -\- xL. At 100 pounds per square inch absolute pressure, 
this is 298.3 + 888.0 X in B.t.u. per pound. Since the heat in 
the steam per pound in the calorimeter is obviously the same as 
before it entered the instrument, we can equate as follows: 

298.3 + 888.0 X = 1 164.8 
X = 0.976 

or the steam is 2.4 per cent wet. 

Barrus Throttling Calorimeter. This is an important varia- 
tion from the type of throttling calorimeter shown in Fig. 14 and 
has been quite widely introduced by Mr. George H. Barrus. In 
this apparatus the temperature of the steam admitted to the 
calorimeter is observed instead of the pressure and a very free 
exhaust is provided so that the pressure in the calorimeter is 
atmospheric. This arrangement simplifies very much the obser- 
vations to be taken, as the quahty of the steam Xi can be calcu- 
lated by equation (65) by observing only the two temperatures 
ti and tc, taken respectively on the high and low pressure sides of 
the orifice in the calorimeter. This calorimeter is illustrated in 
Fig. 15. The two thermometers required are shown in the 
figure. Arrows indicate the path of the steam. f 

The orifice in such calorimeters is usually made about -^^ i^^h 
in diameter, and for this size of orifice the weight of steam { 

* Marks and Davis' Steam Tables and Diagrams (ist ed.), page 24. 
t Transactions of American Society of Mechanical Engineers, vol. XI, page 790. 
X In boiler tests corrections should be made for the steam discharged from the 
steam calorimeters. 



PROPERTIES OF STEAM 



79 



Connection for 
Manometer 




Fig. 14. — Simple Throttling Steam Calorimeter. 



i k I 

//, /' ooooo<>oo6ooo o lidrii 





Fig. 15. — Barrus' Throttling Steam Calorimeter. 



discharged per hour at 175 pounds per square inch absolute 
pressure is about 60 pounds. It is important that the orifice 
should always be kept clean, because if it becomes obstructed 
there will be a reduced quantity of steam passing through the 



8o ENGINEERING THERMODYNAMICS 

instrument, making the error due to radiation relatively more 
important. 

In order to free the orifice from dirt or other obstructions the 
connecting pipe to be used for attaching the calorimeter to the 
main steam pipe should be blown out thoroughly with steam 
before the calorimeter is put in place. The connecting pipe 
and valve should be covered with hair felting not less than 
I inch thick. It is desirable also that there should be no leak 
at any point about the apparatus, either in the stuf&ng-box of 
the supply valve, the pipe joints or the union. 

With the help of a diagram* giving the quahty of steam directly 
the Barrus calorimeter is particularly well suited for use in 
power plants, where the quality of the steam is entered regu- 
larly on the log sheets. The percentage of moisture is obtained 
immediately from two observations without any calculations. 

Separating Calorimeters. It was explained on page 76 that 
throttling calorimeters cannot be used for the determination of 
the quality of steam when for comparatively low pressures the 
moisture is in excess of 2 per cent, and when for average boiler 
pressures in modern engineering practice it exceeds 5 per cent. 
For higher percentages of moisture than these low limits sepa- 
rating calorimeters are most generally used. In these instru- 
ments the water is removed from the sample of steam by mechan- 
ical separation just as it is done in the ordinary steam separator 
installed in the steam mains of a power plant. There is pro- 
vided, of course, a device for determining, while the calorimeter 
is in operation, usually by means of a calibrating gage glass, the 
amount of moisture collected. This mechanical separation de- 
pends for its action on changing very abruptly the direction of 
flow and reducing the velocity of the wet steam. Then, since 
the moisture (water) is nearly 300 times as heavy as steam at 
the usual pressures delivered to the engine, the moisture will be 
deposited because of its greater inertia. 

When a U-tube manometer is used to determine the pressure 
in a calorimeter of the type illustrated in Fig. 14, this pressure 
* See page 117, and also Moyer's Power Plant Testing, 2d edition, pages 58-60. 



PROPERTIES OF STEAM 



8l 



can be obtained very accurately, and an excellent means is pro- 
vided for calibrating the thermometer in the calorimeter just as 
it is to be used. The calibration would be made, of course, by 
the method of comparing with the temperature corresponding to 
known pressures. In order to avoid having superheated steam 
in the calorimeter for this calibration the felt or similar material 
usually needed for covering the valves and nipples between the 
main steam pipe and the calorimeter 
should be kept saturated with cold 
water. 

Fig. 1 6 illustrates a form of 
separating calorimeter having a 
steam jacketing space which re- 
ceives Hve steam at the same tem- 
perature as the sample. Steam is 
supplied through a pipe A, dis- 
charging into a cup B. Here the 
direction of the flow is changed 
through nearly i8o degrees, causing 
the moisture to be thrown outward 
through the meshes in the cup into 
the vessel V. The dry steam passes 
upward through the spaces between 
the webs W, into the top of the- 
outside jacketing chamber J, and 

is finally discharged from the Fig. i 6. — Separating Calorimeter. 

bottom of this steam jacket through 

the nozzle N. This nozzle is considerably smaller than 
any other section through which the steam flows, so that there 
is no appreciable difference between the pressures in the calo- 
rimeter proper and the jacket. The scale opposite the gage 
glass G is graduated to show in hundredths of a pound at the 
temperature corresponding to steam at ordinary working pres- 
sures, the variation of the level of the water accumulating. A 
steam pressure gage P indicates the pressure in the jacket J, 
and since the flow of steam through the nozzle N is roughly 




82 ENGINEERING THERMODYNAMICS 

proportional to the pressure, another scale in addition to the 
one reading pressures is provided at the outer edge of the dial. 
A petcock C is used for draining the water from the instru- 
ment, and by weighing the water collected corresponding to a 
given difference in the level in the gage G the graduated scale 
can be readily calibrated. Too much reliance should not be 
placed on the readings for the flow of steam as indicated by the 
gage P unless it is frequently calibrated. Usually it is very 
little trouble to connect a tube to the nozzle N and condense 
the steam discharged in a large pail nearly filled with water. 
When a test for quality is to be made by this method the pail 
nearly filled with cold water is carefully weighed, and then at 
the moment when the level of the water in the water gage G has 
been observed the tube attached to the nozzle N is immedi- 
ately placed under the surface of the water in the pail. The 
test should be stopped before the water gets so hot that some 
weight is lost by '' steaming." The gage P is generally cali- 
brated to read pounds of steam flowing in ten minutes. For the 
best accuracy it is desirable to use a pail with a tightly fitting 
cover into which a hole just the size of the tube has been cut. 

If W is the weight of dry steam flowing through the orifice N 
and w is the weight of moisture separated, the quality of the 
steam is 

(66) 



W -\-w 



Condensing or Barrel Calorimeter. For steam having a large 
percentage of moisture (over 5 per cent) the condensing or 
barrel calorimeter will give fairly good results if properly used. 
In its simplest form it consists of a barrel placed on a platform 
scale and containing a known weight of cold water. The steam 
is introduced by a pipe reaching nearly to the bottom of the 
barrel. The condensation of the steam raises the temperature 
of the water, the loss of heat by the wet steam being equal to 
the gain of heat by the cold water. It will, therefore, be neces- 
sary to observe the initial and final weights, the initial and 



PROPERTIES OF STEAM 83 

final temperatures of the water in the barrel and the temperature 
of the steam. 

Let W = original weight of cold water, pounds. 
w = weight of wet steam introduced, pounds. 
ti = temperature of cold water, degrees Fahrenheit. 
4 = temperature of water after introducing steam. 
ts = temperature of steam. 
L = latent heat of steam at temperature ts. 
X = quahty of steam. 
Then, 

Heat lost by wet steam = heat gained by water. 

wxL + w{ts-h) =W {h- k) 

^ ^ W{k-k) -w{ts-h) ^^^^ 

The accuracy of the results depends obviously upon the accu- 
racy of the observation, upon thorough stirring of the water so 
that a uniform temperature is obtained and upon the length of 
time required. The time should be just long enough to obtain 
accurate differences in weights and temperatures; otherwise, 
losses by radiation will make the results much too low. 

Equivalent Evaporation and Factor of Evaporation. For the 
comparison of the total amounts of heat used for generating 
steam (saturated or superheated) under unlike conditions it is 
necessary to take into account the temperature k at which the 
water is put into the boiler as well as also the pressure P at 
which the steam is formed.* These data are of much impor- 
tance in comparing the results of steam boiler tests. The basis 
of this comparison is the condition of water initially at the boil- 
ing point for ''atmospheric" pressure or at 14.7 pounds per 
square inch; that is, at 212° F. and with steaming taking place 
at the same temperature. For this scandard condition, then, 

* As the pressure P increases the total heat of the steam also increases; but 
as the initial temperature of the water ("feed temperature") increases the value 
of the heat of the hquid decreases. 



84 ENGINEERING THERMODYNAMICS 

h = o and H = L = 970.4 B.t.u. per pound. Evaporation 
under these conditions is described as, 

" from (a feed- water temperature of) and at (a pressure corre- 
sponding to the temperature of) 212° F." 

To illustrate the application of a comparison with this stand- 
ard condition let it be required to compare it with the amount 
of heat required to generate steam at a pressure of 200 pounds 
per square inch absolute with the temperature of the water sup- 
plied (feed water) at 190° F. 

For P = 200 pounds per square inch absolute the heat of the 
liquid h = 354.9 B.t.u. per pound and the heat of evaporation 
(L) is 843.2 B.t.u. per pound. For t = 190° F. the heat of the 
liquid (ho) is 157.9 B.t.u. per pound. The total heat actually- 
required in generating steam at these conditions is, therefore, 

843.2 + (354.9 — 157.9) = 1040.2 B.t.u. per pound. 

The ratio of the total heat actually used for evaporation to 
that necessary for the condition defined by '' from and at 212° F.'^ 
is called the factor of evaporation. In this case it is the value 

1040.2 -7- 970.4 = 1.07. 

If we write F for factor of evaporation, h and L respectively 
the heats of the liquid and of evaporation corresponding to the 
steam pressure and ho the heat of the liquid corresponding to the 
temperature of feed water, then 

p ^ L + Qi-h) 
970.4 

The actual evaporation of a boiler (expressed usually in pounds 
of steam per hour) multiplied by the factor of evaporation is 
called the equivalent evaporation. 

PROBLEMS 

1. Dry and saturated steam has a pressure of 100 lbs. per sq. in. abso- 
lute. What is the temperature of the steam ? ^wj. 327.8° F. 

2. What is the volume per pound of this steam ? ^;z5. 4.429 cu. ft. 

3. What is the heat of the liquid per pound of this steam? 

Ans. 298.3 B.t.u. 



PROPERTIES OF STEAM 85 

4. What is the latent heat per pound of this steam ? 

Ans. 888.0 B.t.u. 

5. What is the total heat above 32° F. per pound of this steam? 

Ans. 1 186.3 B.t.u 

6. Dry and saturated steam has a temperature of 300° F. What is 
its pressure? Ans. 67.0 lbs. per sq. in. absolute. 

7. How many British thermal units would be required to raise the tem- 
perature of I lb. of water from 32 degrees to the boiling point as stated in 
problem 6? Ans. 269.6 B.t.u. 

8. How many British thermal units are required to evaporate i lb. of 
this water into dry and saturated steam under the conditions of problem 6 ? 

Ans. 909.5 B.t.u. 

9. How many British thermal units are required to generate i lb. of 
dry and saturated steam from water at 32° F. under the conditions of 
problem 6? Ans. 11 79.1 B.t.u. 

10. A closed tank contains 9 cu. ft. of dry and saturated steam at a 
pressure of 150 lbs. per sq. in. absolute. 

(a) What is its temperature? Ans. 358.5** F. 

(b) How many pounds of steam does the tank contain? 

Ans. 2.988 lbs. 

11. A boiler generates dry and saturated steam under a pressure of 200 
lbs. per sq. in. absolute. The feed water enters the boiler at 60° F. 

(a) What is the temperature of the steam? Ans. 381.9° F. 

(b) How many British thermal units are required to generate 

I lb. of this steam if this feed water is admitted at 32° F.? 

Ans. 1198.1 B.t.u. 

(c) How many British thermal units are required to raise the 

temperature of i lb. of water from 32° to 60° F.? 

Ans. 28.08 B.t.u. 

(d) How many British thermal units are required to generate 

I lb. of this steam from feed water at 60 degrees into the 
steam at the pressure stated at the beginning of this prob- 
lem? Ans. 1170.02 B.t.u. 

12. One pound of dry and saturated steam is under a pressure of 250 
lbs. per sq. in. absolute. 

(a) What is its internal energy of evaporation ? 

Ans. 742.0 B.t.u. 

(b) What is its total internal energy above 32° F. ? 

Ans. 1116.4 B.t.u. 

(c) How much external work was done during its formation from 

32° F.? * Ans. 85.1 B.t.u. 

(d) How much external work was done during the evaporation? 

Ans. 84.3 B.t.u. 



86 ENGINEERING THERMODYNAMICS 

(e) How much external work was done during the change in 
temperature of the water from 32 degrees to the boiling 
point corresponding to the pressure? Ans. 0.8 B.t.u. 

13. Dry and saturated steam is generated in a boiler and has a tempera- 
ture of 400° F. The feed water enters the boiler at 200° F. 

(a) What pressure is carried in the boiler ? 

Ans. 247.1 lbs. per sq. in. absolute. 

(b) What is the total heat supplied to generate i lb. of this steam? 

A7ZS. 1033.36 B.t.u. 

(c) How much external work was done during its formation? 

Ans. 84.97 B.t.u. 

(d) How much heat was used in increasing the internal energy? 

Ans. 948.39 B.t.u. 
Check this by (^2 — ^1 + II) noting that this assumes no 
external work done in the heating of the liquid. 

Ans. 949.16 B.t.u. 

14. One pound of steam at a pressure of 100 lbs. per sq. in. absolute has 
a quality of 90 per cent dry. What is its temperature? Ans. 327.8° F. 

15. How many British thermal units would be required to raise the 
pound of steam in the above problem from 32° F. to the boiling point corre- 
sponding to the pressure stated? Ans. 298.3 B.t.u. 

16. What would be the volume of a pound of steam for the conditions 
stated in problem 14? Ans. 3.99 cu. ft. 

17. How many heat units (latent heat) are required to evaporate the 
steam in problem 14? Ans. 799.2 B.t.u. 

18. What is the amount of the total heat (above 32° F.) of the steam in 
problem 14? Ans. 1097.5 B.t.u. 

19. What would be the external work of evaporation of the steam in 
problem 14? Ans. 73.26 B.t.u. 

20. How much external work (above 32° F.) is done in making steam as 
in problem 14? Ans. 73.56 B.t.u. 

21. What is the internal energy of evaporation of the steam in prob- 
lem 14? Ans. 725.9 B.t.u. 

22. What is the total internal energy of the steam in problem 14? 

Ans. 1023.94 B.t.u. 

23. A tank contains 9 cu. ft. of steam at 100 lbs. per sq. in. absolute 
pressure which has a quality of 95 per cent. How many pounds of steam 
does the tank contain ? ' Ans. 2.14 lbs. 

24. Two pounds of steam have a volume of 8 cu. ft. at a pressure of 100 
lbs. per sq. in. absolute. What is the quality? Ans. 90.5 per cent dry. 

Calculate its total heat above 32° F. Ans. 2003.8 B.t.u. 

25. One pound of steam having a quahty of 0.95 has a temperature of 
325° F. What is the pressure? Ans. 96.15 lbs. per sq. in. absolute. 



PROPERTIES OF STEAM 87 

26. One pound of steam at a pressure 225 lbs. per sq. in. absolute has a 
temperature of 441.9° F. Is it superheated or saturated? How many 
degrees of superheat has it? Ans. 50° F. sup. 

What is the total heat required to generate such steam from water at 
32° F.? Ans. 1232.7 B.t.u. 

What is its volume? Ans. 2.23 cu. ft. 

How much external work was done (above 32° F.) in generating it? 

Ans. 92.2 B.t.u. 
How much internal energ>^ above 32° F. does it contain? 

Ans. 1 140.5 B.t.u. 

27. One pound of steam at a pressure of 300 lbs. per sq. in. absolute has 
a volume of 1.80 cu. ft. 

Is it saturated or superheated? 

What is its temperature? Ans. 507.5° F. 

How much superheat has it? Ans. 90° F. 

How much is its total heat above 32° F.? -Ans. 1262.8 B.t.u. 

What is its total internal energy ? Ans. 1163.8 B.t.u. 

28. Steam in a steam pipe has pressure of 110.3 lbs. per sq. in. by the 
gage. A thermometer in the steam registers 385° F. Atmospheric pres- 
sure is 14.7 lbs. per sq. in. absolute. Is the steam superheated, and if 
superheated how many degrees ? 

29. Steam at a pressure of 200 lbs. per sq. in. absolute passes through 
a throttling calorimeter. After expansion into the calorimeter the tem- 
perature of this steam is 250° F. and the pressure 15 lbs. per sq. in. abso- 
lute. What is the quality ? Ans. 0.965. 

30. Steam at a temperature of 325° F. passes through a throttling 
calorimeter. In the calorimeter the steam has a pressure of 16 lbs. per 
sq. in. absolute and a temperature of 236.3° F. What is the quality? 

Ans. 0.973. 

31. Steam at 150 lbs. per sq. in. absolute pressure passes through a 
throttling calorimeter. Assuming that the lowest conditions in the calorim- 
eter for measuring the quality is 10° F. superheat and the pressure in the 
calorimeter is 15 lbs. per sq. in. absolute, what is the largest percentage of 
wetness the calorimeter is capable of measuring under the above conditions ? 

Ans. 4.3 per cent wet or a quality of 0.957. 

32. In a ten-minute test of a separating calorimeter the quantity of dry 
steam passing through the orifice is 9 lbs. The quantity of water sepa- 
rated was I lb. What was the quality? Ans. 0.90. 

33. A barrel contains 400 lbs. of water at a temperature of 50° F. Into 
this water steam at a pressure of 125 lbs. per sq. in. absolute is admitted 
until the temperature of the .water and condensed steam in the barrel 
reaches a temperature of 100° F. The weight of the water in the barrel 
was then 418.5 lbs. What was the quality? Ans. 0.958. 



CHAPTER VI 



PRACTICAL APPLICATIONS OF THERMODYNAMICS TO 
THERMAL MACHINERY 

Refrigerating Machines or Heat Pumps. By a refrigerating 
machine or heat pump is meant a machine which will carry heat 
from a cold to a hotter body.* This, as the second law of 
thermodynamics asserts, cannot be done by a self-acting proc- 
ess, but it can be done by the expenditure of mechanical work. 
Any heat engine will serve as a heat pump if it be forced to 
trace its indicator diagram backward, so that the area of the 




Volume 
Fig. 17. — Pressure- volume Diagram of Camot Cycle. 

diagram represents work spent on, instead of done by, the 
working substance. Heat is then taken in from the cold body 
and heat is rejected to the hot body. 

Take, for instance, the Carnot cycle, using air as working sub- 
stance (Fig. 17), and let the cycle be performed in the order 
dcba, so that the area of the diagram is negative, and repre- 
sents work spent upon the machine. In the stage dc, which is 

* This statement is not at variance with our knowledge that heat does not 
flow of itself from a cold body to a hotter body. 

88 



PRACTICAL APPLICATIONS OF THERMODYNAMICS 89 

isothermal expansion in contact with the cold body R (as in Fig. 
9, page 45), the air takes in a quantity of heat from R equal to 
MRT2 loge r (equation (29)), and in stage ba it gives out to the 
hot body H a quantity of heat equal to MRTi loge r. There is 
no transfer of heat in stages cb and ad. Thus R, the cold body, 
is constantly being drawn upon for heat and can therefore be 
maintained at a temperature lower than its surroundings. In an 
actual refrigerating machine operating with air, of the kind that 
might be used for making ice, the cold body R consists of a coil 
of pipe through which brine circulates while " working " air is 
brought into contact with the outside of the pipe. The brine is 
kept, by the action of the machine, at a temperature below 32° F. 
and is used in its turn to extract heat by conduction from the 
water which is to be frozen to make ice. The ''cooler" H, which 
is the relatively hot body, is kept at as low a temperature as 
possible by means of circulating water, which absorbs the heat re- 
jected to H by the " working " air. 

The use of a regenerator, as in Stirling's engine (page 55), 
may be resorted to in place of the two adiabatic stages in the 
Carnot cycle just explained with .the advantage of making the 
machine much less bulky. Refrigerating machines of this kind 
using air as working substance, with a regenerator, were intro- 
duced by Dr. A. C. Kirk and have been widely used.* The 
working air is completely enclosed, which allows it to be in a 
compressed state throughout, so that even its lowest pressure 
is much above that of the atmosphere. This makes a greater 
mass of air pass through the cycle in each revolution of the 
machine, and hence increases the performance of a machine of 
given size. 

In another class of refrigerating machines the working sub- 
stance, instead of being air, consists of a Hquid and its vapor, 
and the action proceeds by alternate evaporation under a low 
pressure and condensation under a relatively high pressure. A 

* See Kirk, On the Mechanical Production of Cold, Proc. Inst, of C. E., vol. 
XXXVII, 1874. Also lectures on Heat and its Mechanical Applications, in the 
same proceedings for 1884. 



go ENGINEERING THERMODYNAMICS 

liquid must be chosen which evaporates at the lower extreme of 
temperature under a pressure which is not so low as to make 
the bulk of the engine excessive. Ammonia, ether, sulphurous 
acid, and other volatile liquids have been used. Ether machines 
are inconveniently bulky and cannot be used to produce intense 
cold, for the pressure of that vapor is only about 1.3 pounds per 
square inch at 4° F., and to make it evaporate at any tempera- 
ture nearly as low as this would require the cylinder to be exces- 
sively large in proportion to the performance. This would not 
only make the machine clumsy and costly, but would involve 
much waste of power in mechanical friction. The tendency of 
the air outside to leak into the machine is another practical ob- 
jection to the use of so low a pressure. With ammonia a dis- 
tinctly lower limit of temperature is practicable: the pressures 
are rather high and the apparatus is compact. 

The standard systems of mechanical refrigeration are : * 

(A) The dense-air system, so-called because the air which is 
the medium is never allowed to fall to atmospheric pressure, so 
as to reduce the size of the cylinders and pipes through which a 
given weight is circulating. 

(B) The compression system, using ammonia, carbon dioxide 
or sulphur dioxide, and so-called to distinguish it from the third 
system, because a compressor is used to raise the pressure of the 
vapor and deliver it to the condenser after removing it from the 
evaporator. 

(C) The absorption system, using ammonia, and so-called be- 
cause a weak water solution removes vapor from the evaporator 
by absorption, the richer aqua ammonia so formed being pumped 
into a high-pressure chamber called a generator in communica- 
tion with the condenser, where the ammonia is discharged from 
the liquid solution to the condenser by heating the generator, 
to which the solution is delivered by the pump. 

No matter what system is used, circulating water is employed 
to receive the heat, the temperature of which Kmits the highest 

* Lucke's Engineering Thermodynamics, page 1148. 



PRACTICAL APPLICATIONS OF THERMODYNAMICS g'l 

temperature allowable in the system and indirectly the highest 

pressure. 

The dense or closed air system is illustrated in Fig. i8, in 



A 









1 


Circulatin 


^ T 










1 Water 1 
1 1 




A 






Water Cooler 














■^ 




Cool Com- 








Hot Com- 


pressed Air 




pressed Air 


\f j\ 


















^ 










Expansion _ 
Cylinder 




Compressor 
Cylinder 




E 
Engine 
Cylinder 




















^ 


V 
















\ 








Cold Low 




Warmed Low 


Pressure Air 








Pressure Air 
















'-' 


■ ^ 


\ 




^ 










1 ^ B . 
















1 " 






Brine Tank '' 






1 "^ 






1 


-^ 





jCirculating^ Brmej 
Warm i Pipes \!/Cold 

Brine Brine 

I : 

Fig. 1 8. — Dense Air System of Refrigeration. 

which air previously dried of moisture is continuously circulated. 
The engine cylinder, E, furnishes power * to drive the compressor 

* Since the work done by the expansion of the cool-compressed air is less than 
that necessary for the compressing of the air taken from the brine coils through 
the same pressure conditions, a means must be employed to make up for the dif- 
ference, and for this purpose the engine cylinder is used. 



c 


^ M 


B 








■1 




k 




D 






w 


\A 








N 





92 ENGINEERING THERMODYNAMICS 

cylinder, F. This cylinder delivers hot-compressed air into a 
cooler, A, where it is cooled, and then passed on to the expan- 
sion cyHnder, G (tandem-connected to both, the compressor, 
F, and to the engine cyHnder, E), which in turn sends cold low- 
pressure air first through the refrigerating coils in the brine 
tank, B, and then back to the compressor cyHnder, F; thus 
the air cycle is completed. The courses of the circulating water 
and also of the brine are shown by the 
dotted Hues. 

The dense-air cycle in a pressure- 
volume diagram is represented in Fig. 19, 
in which BC is the deHvered volume of 
hot-compressed air; CM is the volume of 

Volume , T . 1.1-1 

Fig. 19.— Pressure- volume cooled air admitted in the expansion 
Diagram of Dense Air cyHnder; MB the reduction in volume 
yceo engera ion. ^^^ ^^ ^^ water cooler; MN, the expan- 
sion; NA the refrigeration or heating of the air by the brine, 
and AB the compression. This operation is but a reproduction 
of that previously described. 

The compression system for ammonia or similar condensable 
vapors is shown in Fig. 20. The figure only illustrates the 
essential members of a complete compression refrigerating sys- 
tem. B represents the direct-expansion coil in which the work- 
ing medium is evaporated; F, the compressor or pump for 
increasing the pressure of the gasified ammonia; E, the engine 
cyHnder, — the source of power; W, the condenser for cooling 
and Hquefying the gasified ammonia; and V a throttHng valve 
by which the flow of liquefied ammonia under the condenser 
pressure is controlled as it flows from the receiver R to the ex- 
pansion coils; B (the brine tank), in which a materiaUy lower 
pressure is maintained by the pump or compressor in order 
that the working medium may boil at a sufficiently low tem- 
perature to take heat from and consequently refrigerate the 
brine which is already cooled. 

These descriptions of the refrigeration systems will serve as a 
foundation for a general understanding of refrigerating. 



PRACTICAL APPLICATIONS OF THERMODYNAMICS 



93 



COEFFICIENT OF PERFORMANCE OF REFRIGERATING MACHINES 

Heat extracted from the cold body 
Work expended 

This ratio may be taken as a coefficient of performance in esti- 
mating the merits of a refrigerating machine from the thermo- 

I i 

I Circulating I 

Y Water A 

! I 



High Pres-^ ' 
sure Liquid 



W 
Condenser 



R 

Ammonia 
Receiver 



High Pres- 
> ^sure Vapor 



CompreBBor 
Cylinder 



E 
Engine 
Cylinder 



^El 



Throttling or 
Expansion Valve 

Low Pressure Liquid 



a 



Brine ^ Tank 



Low Pres- 
sure Vapor 



l« — Circulating Brine — s-l 

I Pipes i 

Warm | I Cold 

Brine I | Brine 

Fig. 20. — Compression System of Refrigeration. 

dynamic point of view. When the limits of temperature Ti 
and T2 are assigned it is very easy to show by a sHght variation 
of the argument used in Chapter IV that no refrigerating machine 



-94 ENGINEERING THERMODYNAMICS 

can have a higher coefficient of performance than one which is 
reversible according to the Carnot method. For let a refriger- 
ating machine S be driven by another R which is reversible 
and is used as a heat-engine in driving S. Then if S had a higher 
coefficient of performance than R it would take from the cold 
body more heat than R (working reversed) rejects to the cold 
body, and hence the double machine, although purely self-actingj 
would go on extracting heat from the cold body in violation of 
the Second Law (page 3). Reversibility, then, is the test of 
perfection in a refrigerating machine just as it is in a heat-engine. 
When a reversible refrigerating machine takes in all its heat, 
namely Qc at T2 and rejects all, namely Qa at Ti and if we repre- 
sent the heat equivalent of the work done by PT = Q^ ~ Qc, then 
the coefficient of performance is as already defined, 

Qc _ Qc T^ 

W Qa-Qc T,-T2 

Hence — and the inference is highly important in practice — 
the smaller the range of temperature, the better is the perform- 
ance. To cool a large mass of any substance through a few 
degrees will require much less expenditure of energy than to 
cool one-fifth of the mass through five times as many degrees, 
although the amount of heat extracted is the same in both cases. 
If we wish to cool a large quantity, say of water 1 or of air, it is 
better to do it by the direct action of a refrigerating engine 
working through the desired range of temperature, than to cool 
a portion through a wider range and then let this mix with the 
rest. This is only another instance of a wide, general principle, 
of which we have had examples before, that any mixture or 
contact of substances at different temperatures is thermody- 
namically wasteful because the interchange of heat between 
them is irreversible. Ah ice-making machine, for example, 
should have its lower limit of temperature only so much lower 
than 32° F. as will allow heat to be conducted to the working 
fluid with sufficient rapidity from the water that is to be 
frozen. 



PRACTICAL APPLICATIONS OF THERMODYNAMICS 95 

COMPRESSED AIR 

Air when compressed may be used as the working medium in 
an engine, in exactly the same way as steam. Furthermore, it 
is an agent for the transmission of power and can be distributed 
very easily from a central station for the purpose of driving 
engines, operating quarry drills and various other pneumatic 
tools. The type of machine used for the compression of air is 
that known as a piston-compressor, and consists of a cylinder 
provided with valves and within which there is a reciprocating 
piston. 

Inasmuch as the work performed in the air-cylinder of a com- 
pressor depends on so many variable conditions, it can only be 
studied successfully from an indicator diagram. Imagine in such 
a compressor, that the compression is performed very slowly 
in a conducting cyhnder, so that the air within may lose heat 
by conduction to the atmosphere as fast as heat is generated by 
compression; the process will in that case be isothermal, at the 
temperature of the atmosphere. Imagine further that the com- 
pressed air is distributed to be used in compressed air motors * 
or engines without a change of temperature, and that the 
process of expansion in the compressed air motors or engines is 
also indefinitely slow and consequently isothermal. In that 
case (if we neglect the losses caused by friction in pipes) there 
could be no waste of power in the whole process of transmission. 
The indicator diagram would then be the same per pound of air 
in the compressor as in the air motor or engine, although the 
course of the cycle would be the reverse — that is, it would re- 
trace itself. 

Imagine, on the other hand, that compression and expansion 
are both adiabatic — a state of things which would be approxi- 
mately true, if expansion and compression were performed very 
quickly. The diagram of the compression, Fig. 21, is FCBE 
and that of the air engine as in Fig. 22, is EADF, and, therefore, 

* Compressed air motors are really engines just like steam engines but use 
compressed air instead of steam. 



96 



ENGINEERING THERMODYNAMICS 



CB and AD are both adiabatic lines. The change of volume 
of the compressed air from that of EB to EA occurs through its 
cooling in the distributing pipes, from the temperature pro- 
duced by adiabatic compression down to the temperature of 
the atmosphere. Suppose both diagrams of compressor and of 




Volume 
Fig. 21. — Diagram of Compressor. 



Volume 
Fig. 2 2. — Diagram of Air Engine. 



air engine be superimposed as in Fig. 23, and then sketch an 
imaginary isothermal line between the points A and C, both of 
which are at atmospheric conditions as regards temperature. 

This simple sketch shows that the use of adiabatic compres- 
sion causes a waste of power which is measured by the area 




Volume 
Fig. 23. — Superimposed Diagrams of Figs. 21 and 22. 

ABC, while the use of the adiabatic expansion in the air engine 
involves a further waste, shown by the area ACD. 

In practice the compression cannot be made strictly isother- 
mal for want of time, — ■ the operation of the piston would be 
too slow for practice. The difference between isothermal and 
adiabatic compression (and expansion) can be very clearly shown 
graphically as in Figs. 24 and 25. In this illustration the termi- 
nal points are correctly placed for a certain ratio for both com- 
pression and expansion. Note that in the compressing diagram 



PRACTICAL APPLICATIONS OF THERMODYNAMICS 



97 



(Fig. 24), the area between the two curves ABC represents the 
work lost in compressing due to heating, and the area between 
the two curves, ACMNF (in Fig. 25), shows the work lost by 
cooling during the expansion. The isothermal curve AC will 
be the same for both cases. Illustrations of this sort show the 





Volume 
Fig. 24. — Compression Diagram. 



Volume 
Fig. 25. — Expansion Diagram. 



effect of reheating before expansion, cooling before compression, 
heating during expansion, etc. 

The temperature of the air is prevented as far as possible 
from rising during the compression by injecting water into the 
compressing cylinder, and in this way both the isothermal and 
adiabatic curves will change. The curves which would have 
been PV = a. constant, if isothermal and PV^"^ = a constant, if 
adiabatic will be very much modified. In perfectly adiabatic 
conditions the exponent ^'w" = 1.40 for air, but in practice the 
compressor cylinders are water-jacketed, and thereby part of 
the heat of compression is conducted away, so that ^'w" becomes 
less than 1.40. This value of "w" varies with conditions; gen- 
erally the value is taken as 1.2. 

The problem of economy, obviously, becomes one of abstract- 
ing the heat generated in the air during the process of compres- 
sion. As previously mentioned, this is partially accomplished by 
water-jacketing the cylinders, and also by water injection. 
Nevertheless, owing to the short interval within which the com- 
pression takes place, and the comparatively small volume of 
air actually in contact with the cylinder walls, very little 
really occurs. The practical impossibility of proper cooling to 
prevent waste of energy' leads to the alternative of discharging 
air from one cylinder after partial compression has been effected, 



98 



ENGINEERING THERMODYNAMICS 



into a so-called inter-cooler, intended for removing the heat 
generated during the first compression, and then compressing 
the air to the final pressure in another cyHnder. This opera- 
tion is termed " two-stage " compression and when repeated 
one or more times for high pressures, the term " multi-stage " 
compression applies. 

Referring to Fig. 26 and assuming the compression in a two- 
stage compressor to be adiabatic for each cylinder, the compres- 




Volume 
Fig. 26. — Indicator Diagram of Two-stage Air Compressor. 

sion curve is represented by the broken line ABDE ; the compres- 
sion proceeds adiabatically in the first or low-pressure cyHnder 
to B ; the air is then taken to a cooler and cooled under practi- 
cally constant pressure until its initial temperature is almost 
reached, and its volume reduced from HB to HD; it is then 
introduced to the second or high-pressure cylinder and com- 
pressed adiabatically along the line DE to the final pressure 
condition that was desired. It is seen that the compression 
curve approaches the isothermal line FA.* The isothermal con- 
dition is ob\dously desired and, in consequence, air-machines 
are built to approach that condition as nearly as possible. 

Numerous devices may be applied by the engineer to make 
the expansion curves in his air engines approximate more nearly 
to the isothermal Kne; that is, he may use a preheater or inject 
hot water; or use a compound engine, allowing the air time to 
take up enough heat to restore it more or less nearly to atmos- 
pheric temperature between one stage of expansion and the next. 
By these means the efficiency of the transmitting system as a 
whole (neglecting all losses due to friction in the distributing 

* The line FE represents further cooUng. 



PRACTICAL APPLICATIONS OF THERMODYNAMICS 99 

pipes, in the valves of the engines, etc.), may be considerably 
increased. 

There is, however, another point to be considered. If the 
temperature be allowed to fall materially during expansion, the 
expanding air tends to deposit dew or even snow. To prevent 
this the practice is often followed of passing the compressed air 
through a stove or '^preheater" in order to raise its temperature 
just before it is allowed to expand and so prevent the deposit of 
frozen moisture. When preheaters are used the extra heat 
which they supply is, of course, itself partly converted into 
work.* 

PROBLEMS 

1. If 200 cu. ft. of free air per minute (sea-level) is compressed isother- 
mally and then delivered into a receiver, the internal pressure of which is 
102.9 lbs. per sq. in. absolute, find the theoretical horse-power required. 

Ans. 24,93 h.p. 

2. What will be the net work in foot-pounds per stroke by an air com- 
pressor displacing 3 cu. ft. per stroke, compressing air from an atmospheric 
pressure of 15 lbs. per sq. in. absolute, to a gage pressure of 75 lbs.? (Iso- 
thermal.) Ans. 11,595. 

3. What horse-power will be needed to compress adiabatically 1500 
cu. ft. of free air per minute to a gage pressure of 58.8 lbs., when n equals 
1.4? Ans. 197 h.p. 

4. A compressed-air motor without clearance takes air at a condition 
of 200 lbs. per sq. in. (gage) and operates under a cut-off at one-fourth 
stroke. What is the work in foot-pounds that can be obtained per cubic 
foot of compressed air, assmning free air pressure of 14.5 lbs. and n equal 
to 1. 41? Ans. 54,936 ft.-lbs. 

5. Find the theoretical horse-power developed by 3 cu. ft. of air per 
minute having a pressure of 200 lbs. per sq. in. absolute, being admitted 
and expanded in an air engine with one-fourth cut-off. The value of n is 
1.2. (Neglect clearance.) Ans. 5.01 h.p. 

6. Compute the net saving in energy that is effected by compressing 
isothermally instead of adiabatically 50 cu. ft. of free air to a pressure of 

* On the subject of transmission of power by compressed air, reference should 
be made to papers by Professor Richards in Bulletin No. 63 of Engineering Experi- 
ment Station of Univ. of 111.; Weymouth on " Problems in Gas Engineering," 
Trans. American Society of Mechanical Engineers, vol. 34 (1912), pages 185-234; 
Baker on "Expansion and Temperature Drop of Compressed Air," Trans. 
A. S. M. E., vol. 2)2)1 pages 918-919. 



lOO ENGINEERING THERMODYNAMICS 

200 lbs. per sq. in. gage. Barometer = 14 lbs. per sq. in., and a tempera- 
ture of 70° F. What is the increase in intrinsic energy during each kind 
of compression? How much heat is lost to the jacket-water during each 
kind of compression ? 

Ans. 144,000 ft.-lbs.; —o isothermal; 299.500 ft. -lbs. adiabatic; 275,500 
ft.-lbs. isothermal; o adiabatic. 

7. Let a volume of 12 cu. ft. of free air be adiabaticaUy compressed in 
one stage from atmospheric pressure (15 lbs.) to 85 lbs. gage; the initial tem- 
perature of the air being 70° F. 

(a) What is the volume and temperature of the air after the com- 

pression? Ans. 3.12 cu. ft.; 920'' F. absolute. 

(b) Suppose this heated and compressed air be cooled to an initial 

temperature of 60° F., what is its pressure for that condi- 
tion? Ans. 56.65 lbs. absolute. 

(c) Now if the air occupies such a voliune as foimd in (a) and at 

an absolute pressure as in (b), at a temperature of 60° F., 
and is then allowed to expand adiabaticaUy down to atmos- 
pheric pressure (15 lbs.), what is the temperature of the 
expanded air in Fahrenheit degrees ? Its volume as well ? 

Ans, —106.72° F.; 8 cu. ft. 



CHAPTER VII 
ENTROPY 

Pressure-volume diagrams are useful for determining the 
work (in foot-pounds), done during a cycle, but they are of very 
limited use in analyzing the heat changes involved. It has, 
therefore, been found desirable to make use of a diagram which 
shows directly by an area the number of heat units (instead of 
foot-pounds) involved during the processes constituting a cycle. 
In order that an area shall represent heat units instead of work 











Tg 


^^ 


C^^ 






ft 


Tx 








-2 




e 


■SH 


/ 



dH 



Fig. 27. — Diagram of Entropy. 



Entropy- (p 
Fig. 28. — Analysis of Entropy Diagram. 



units the coordinates must be such that their product will give 
heat units. If the ordinates are in absolute temperature, the 
abscissas must be heat units per degree of absolute temperature, 

77 77 

that is, — ;, for then T X ~ = H, the amount of heat added 

during the process from A to B (Fig. 27). 

This simple ratio, heat added divided by the absolute tempera- 
ture during addition, can be employed when the temperature re- 
mains constant; but when the temperature changes, a different 
form of expression must be developed. Suppose that the heat 
H is divided up into a number of small increments dH (Fig. 28), 



I02 ENGINEERING THERMODYNAMICS 

and that each small increment of heat is divided by the average 

absolute temperature at which the heat change occurs. We 

dH 
will then have a series of expressions — which, when summed 

up, will give the total change in the abscissas. This quantity 

/ — when multiplied by the average absolute temperature 

between C and D will give the total amount of heat added 
during the process. Mathematically expressed, the change in 
the abscissas is 

d<f>=— ^^ '^^J'Y ^^ 

and the heat change involved is 

dH = Tdcj> or H=CTdcf>, (69) 

The quantity 4> in the equations is known as the increase in en- 
tropy of the substance, and may be defined as a quantity which, 
when multiplied by the average absolute temperature occurring 
during a process, will give the number of heat units (in B.t.u.) 
added or abstracted as heat during the process. The " increase 
in entropy " is employed rather than entropy itself, because we 
are concerned only with the differences in entropy, and further- 
more we could not calculate the absolute entropy, because the 
specific heat of a substance is not accurately known at low 
temperatures. 

This definition of entropy means that in a diagram such as 
Fig. 28, where the ordinates are absolute temperatures, and the 
abscissas are entropies as calculated above some standard tem- 
perature, the area under any line CD gives the number of heat 
units added to the substance in passing from a temperature Ti 
and entropy 0i = Oe, to a temperature T2 and entropy 02 = Of 
(or the number of heat units abstracted in passing from T2 to Ti) . 

Let us apply the above conceptions of entropy to the analysis 
of a Carnot cycle. A pound of the working substance is expanded 
isothermally. On a T-^ (temperature-entropy) diagram (Fig. 




ENTROPY 103 

29), this process would be represented by line AB, where the 
temperature remains constant at Ti, and where the entropy in- 
creases from Oe to Of, because of the addition of heat that is 
required to keep the temperature constant. The amount of 
heat that is added is given by the area ABfe. 

The next process is adiabatic expansion from Ti to T^. Heat 
is neither added to nor abstracted from the substance during 
this expansion. Hence the entropy 
remains constant, as indicated by BC. l^ 

The substance is now isothermally | 
compressed along CD, the temperature, | 
of course, remaining constant at T2 and 1 
the entropy decreasing because of the *^ 
abstraction of heat equal to the area Entropj'-^ 

under CD, i.e., CDef. Fig. 29. — Entropy Diagram of 

The last process of the cycle is ^^^'^ ^^ ^' 

adiabatic compression from D to A, no heat being added or 
abstracted, and the entropy, therefore, remaining constant.* 

To determine the amount of net work done during this cycle 
we can employ the familiar relation 
(Heat equivalent of) work done = heat added — heat rejected. 

(70) 

Applying equation (70) to Fig. 29 we have 

Heat added = ABfe ; heat rejected = CDef. Therefore, 

Work done (in B.t.u.) = ABfe - CDef 

= ABCD. (71) 

Appl)dng the expression for efficiency, namely, 
^ work done 
heat added 
we obtain then, 

„^ . ABCD 

Einciency = -^i 

^ . ABfe 

Now the question may arise as to why the entropy remains 

constant during adiabatic expansion, as from B to C, when it is 

* Adiabatic lines are sometimes called isentropic lines (lines of equal entropy). 



I04 ENGINEERING THERMODYNAMICS 

known that at C there is less intrinsic energy (i.e., less heat) in 
the substance than at B, the decrease being MCv {Ti — T2). 
The answer is that while heat disappeared from the substance 
during the process, it disappeared as work and not as heat. 
The case of constant entropy during adiabatic expansion is thus 
found to be in accord with the explanation of entropy (as given 
on page 102), which states that the heat involved in a change of 
entropy must be added or abstracted as heat, that is, must be 
conducted or radiated to or from the substance. 

From the foregoing discussion two important conclusions may 
be drawn in regard to the use of the T-<f> diagram: 

1. If any Jieat process be represented by a curve on a T-cl> 
diagram, the heat involved during the process is equal to the 
area under the curve, that is, between the curve and the axis of 
absolute temperature. 

2. If a cycle of heat processes be represented on a T-<f) 
diagram by a closed figure, the net work done is equal to the 
enclosed area, that is, the enclosed area measures the amount 
of heat that was converted into work. 

As has been stated previously, T-(f) diagrams are very use- 
ful for analyzing heat processes, and are often referred to as 
^'heat diagrams." They find particularly useful apphcation in 
steam engineering, as indicated by the following: 

1. Graphical analysis of heat transfers in a steam engine 
cylinder (pages 129-133). 

2. Determination of quality of steam during adiabatic ex- 
pansion (page 116). 

3. Quick calculation of efficiency according to Rankine cycle 
(page 129). 

4. Steam turbine calculations (page 123). 

The applications of entropy to heat engineering will be readily 
appreciated as the subject is developed. 

Temperature-Entropy Diagrams for Steam. Practically all 
the diagrams that have been shown in the preceding chapters 
indicated the relations between pressure and volume. Such 
diagrams show very well the action of steam in a reciprocating 



ENTROPY 



105 



o800- 

3 

■§600- 

t^oo- 

H c 
,.• 200 




steam engine, where steam is admitted at the beginning of a 
stroke, expanded and exhausted or rejected from the cylinder 
on the return stroke. Steam turbines operate differently, as 
there is a continual flow of steam into the high pressure end and 
a continual flow of low pressure steam from the exhaust. The 
resulting continual and uninterrupted drop in pressure and tem- 
perature cannot be satisfactorily represented with the pressure- 
volume diagrams that have preceded. 

Another kind of diagram is, therefore, universally used by 
steam engineers. In this diagram, as has been stated, any 
surface represents accurately to given scales, a definite quan- 
tity of heat. Absolute temperatures 
(r) are the ordinates, and the entro- 
pies (</)) are the abscissas. 

Figure 30 shows a simple heat dia- 
gram laid out with absolute tempera- 
ture and entropy for the coordinates. 
Steam at a certain condition of tem- 
perature and entropy is represented 
here by the point A. Then if some 
heat is added, increasing both tem- 
perature and entropy, the final condition is represented by 
the point B, and the area ABCD represents the heat added in 
passing from the condition at A to the condition at B. Such a 
diagram is called a temperature-entropy diagram, although the 
name " heat diagram " is just as appropriate, since every area 
represents a definite amount of heat. 

Another temperature-entropy diagram is shown in Fig. 31 rep- 
resenting by the various shaded areas the heat added to water 
at 32° F. to completely vaporize it at the pressure Pi. The 
unshaded area under the irregular curve AA'B represents the 
heat in a pound of water at the freezing point, 32° F. or 492 de- 
grees in absolute Fahrenheit temperature. A'B represents the 
increase in entropy due to the latent heat of fusion. The area 
OBCD is the heat added to the water to bring it to the tempera- 
ture of vaporization, or in other words this last area represents 



1.0 

Entropy (0) 



2.0 



Fig. 30. — Simple Temperature- 
entropy Diagram of Steam. 



io6 



ENGINEERING THERMODYNAMICS 



the heat of the liquid {h) given in the steam-tables for the pres- 
sure Pi. Further heating after vaporization begins is at the 
constant temperature Ti corresponding to the pressure Pi, and 




Dry Steam or 
Saturation Line 



0.5 1.0 1.5 2.0 

Entropy (0) 

Fig. 31. — Temperature-entropy Diagram of Steam. 



Ord. 


Abs. 


/ 






Eahr.^ 
328- 


SfJ^o- o/Ti s ek 






"1 


139 
32°< 


5007T2 


P \m 




2 








\ 




To 




1 g 






2 






1 13 






3 






! - 












1 -t^ 






c3 






1 '^ 






j^ 






■^ 






K 












a 






-< 






a> 












H 










V 





tt 


6 fc / ^ ^ 


i 


Entropy- 


)ge T2— H 




\OEe\-^, 


Fig. ^ 


52.- 


-Dia 


gram for Calculation of 


Entro 


py of Steam. 



is represented by an increasing area under line CE. When 
'^ steaming " is complete, the latent heat, or the heat of vaporiza- 
tion (L), is the area DCEF. If, after all the water is vaporized, 
more heat is added, the steam becomes superheated, and the 



ENTROPY 107 

additional heat required would be represented by an area to 
the right of E. 

Calculation of Entropy for Steam. In order to lay off the 
increase in entropy as abscissas in the heat diagram for steam, 
it is necessary to develop formulas for calculating </>. For con- 
venience 32° F. has been adopted as the arbitrary starting point 
for calculating increase of entropy, as well as for the other 
thermal properties of steam. Referring to Fig. 32 the entropy 
of water is seen to be o at 32° F. In order to raise the tem- 
perature from To to Ti, h heat units are required, and by the 
equation (68) , the entropy of one pound of the liquid 6 will be 

'^1 dh f V 

(72) 



J To 



T 

or, assuming the specific heat of water to be unity we can write 
'T,dT 



e 



CTidT 
Jto T 



= lOge Ti - loge To = loge -=7' (iS) 

In order to evaporate the water into steam at the boihng 
point Ti, the latent heat, L, must be added at constant tempera- 
ture Ti. The increase of entropy during the "steaming" proc- 
ess is represented by — and is obviously equal to 

Entropy of Evaporation = -rp-- (74) 

Ti 

The total entropy, <f>, of dry saturated steam above 32° F. at 
temperature Ti is, then, 

= « + ^ = log.|l + fi, (75) 

I loll 

and for steam at temperature T2, we have 

* = log.J + ^- 
Iq 1i 

Values of these entropies can be found in nearly all steam-tables. 



io8 



ENGINEERING THERMODYNAMICS 



Isothermal Lines of Steam. When the expansion of steam 
occurs at constant pressure as, for example, in the conversion 
of water into steam in a boiler when the engines are working, we 
have isothermal expansion. It must be obvious from the pre- 
ceding explanation that steam (or any other vapor) can be ex- 
panded or compressed isothermally only when wet. Isothermal 
lines for wet steam, which consists of a mixture of water and its 
vapor, are, therefore, straight lines of uniform pressure. On a 
pressure-volume diagram an isothermal line is consequently 
represented by a horizontal line parallel to the axis of abscissas. 
The horizontal parts of the indicator diagram as illustrated in 




Volume 

Fig. 33. — Indicator Diagram of Ideal Cycle Using Steam. 

Fig* 33 are Knes of constant pressure and, therefore, isothermal 
Knes of steam. On a T-(f) diagram, the isothermal line is repre- 
sented by a line of constant temperature, i.e., by a line parallel 
to the X-axis. 

Adiabatic Lines for Steam. Adiabatic lines will have differ- 
ent curvature as they represent expansion or compression of 
different substances. It will be remembered that the values of 
7 are different for the various gases discussed in preceding 
chapters, and therefore the adiabatic line for each of these gases 
would have a different curvature. In the same way the curva- 
ture of adiabatic lines of steam will vary with the relative amounts 
of steam and water in wet steam. It is worth mentioning here 
that steam which is initially dry, if allowed to expand adiabati- 
cally, will become wet, the percentage of moisture which it will 
contain depending on the extent to which the expansion is car- 



ENTROPY 109 

ried. Also, on any r-0 diagram, an adiabatic (isentropic) line 
is represented by a line parallel to the Y-axis, i.e., by a line of 
constant entropy. If steam is initially wet and is expanded adia- 
batically, it becomes wetter as a rule.* 

In general, in any sort of an expansion in order to keep steam 
at the same relative dryness as it was initially, while it is doing 
work some heat must be suppHed and taken in. And if the ex- 
pansion is adiabatic so that no heat is taken in, a part of the 
steam will be condensed and will form very small particles of 
water suspended in the steam, or it will be condensed as a sort 
of dew upon the surface of the enclosing vessel. 

The relation between pressure and temperature as indicated 
by the steam-tables continues throughout an expansion, pro- 
vided the steam is initially dry and saturated or wet. 

Adiabatic Cixrve for Steam. Whether steam is initially dry 
and saturated or wet, the adiabatic curve may be represented 
by the formula: PV"" = constant. The value of the index n 
depends on the initial dryness of the steam. Zeuner has deter- 
mined the following relation 

oc 

n = LOS"; H 

^^ 10 

Solving this when x = unity (dry and saturated steam) the 
value of n is 1.135, and when x is 0.75, n has the value i.ii.t 

Example. One pound of steam having a quality of 0.95 at a 
pressure of 100 pounds per square inch absolute expands adia- 
batically to 15 pounds per square inch absolute. 

What is the quality at the final condition? 

* When the percentage of water in wet steam is very great and the steam is 
expanded adiabatically there is in many cases a tendency at the beginning of the 
expansion for the steam to become drier. This is very evident from an inspection 
of diagrams like Fig. 32. 

t Rankine gave the value oi n = -\°-, which obviously from the results given 
is much too low if the steam is at all near the dry and saturated condition. His 
value would be about right for the condition when a; = 0.75. In an actual steam 
engine, the expansion of steam has, however, never a close approximation to the 
adiabatic condition, because there is always some heat being transferred to and 
from the steam and the metal of the cyUnder and piston. 



no ENGINEERING THERMODYNAMICS 

Solution. The total entropy at the initial condition equals 
e-Vx- = 0.4743 + 0.95 X 1. 1277 = 1.5456. 

The total entropy at the end of the expansion equals 
e-\-x— = 0.3133 + i.44i6:XJ. 

Since entropy is constant in adiabatic expansion 
0.3133 + 1.4416X = 1.5456, 
from which x = 0.854. 

How much work is done during the expansion? 
Since there is no heat added the work done equals the loss in 
internal energy. 
The internal energy at the end of the expansion equals 
h + xiL = 181.0 + 0.854 X 896.8 = 946.9 B.t.u. 
The internal energy at the initial condition equals 

h + xIl = 298.3 + 0.95 X 806.6 = 1064.6 B.t.u. 
The work equals 

1064.6 — 946.9 = 1 17.7 B.t.u. or 91,576 foot-pounds. 

Example. One pound of steam has a pressure of 100 pounds 
per square inch absolute and a quality of 0.95. It expands 
along amn = I curve to 20 pounds per square inch absolute. 
What is the quality at the end of the expansion? 
Solution. The volume of the steam at the initial condition is 
xXV. 

X X V = 0.95 X 4.429 = 4.207 cubic feet. 
Obviously, PiFi" = P2F2" and since n = i, 

100 X 4-?o7 = 20 X V2, 

V2 = 21.035 cubic feet. 

The volume of dry saturated steam at the end of the expan- 
sion or at 20 pounds per square inch is 20.08. Therefore the 
steam is superheated at end of expansion. How is this known? 



ENTROPY III 

From the superheated steam-tables the amount of superheat cor- 
responding to a specific volume of* 2 1.035 is found to be 29° F. 
What is the work done during the expansion? 

Work = PiFi loge^/ = 100 X 144 X 4-207 loge^^^^ > 
Vi 4.207 

= 100 X 144 X 4.207 X 1.6094 

= 97,499 foot-pounds or 125.3 B.t.u. 

How much heat must be added? 

The energy of the steam at the end of the expansion equals, 

since the pressure is 20 pounds per square inch absolute and 

29 degrees superheat: 

1169.9 ^^ ^ ^^ = 1169.9 — 77.8 = 1092. 1 B.t.u. 

778 

The internal energy of the steam at the initial condition is 

h + xIl = 298.3 + 0.95 X 806.6 = 1064.6. 
The internal energy of the steam was increased during the 
expansion by 1092. i — 1064.6 = 27.5. Therefore 27.5 B.t.u. of 
heat must be suppKed to compensate this increase as well as 
the heat necessary to do the work. The total heat supplied to 
the steam during the expansion would thus be 

27.5 + work of the expansion = 27.5 + 125.3 

= 152.8 B.t.u. 

Example. One pound of steam at a pressure of 150 pounds 
per square inch absolute and a volume of 1.506 cubic feet expands 
under constant pressure until it becomes dry and saturated. 

What is the quahty at the initial condition? 

Solution. The volume of dry and saturated steam at the 
given pressure is 3.012 cubic feet per pound. 

The quality then is -^^ — = o. t;o. 

3.012 

What is the volume of the steam at the final condition? 

The volume a pound then is 3.012 cubic feet since the steam 
is dry and saturated. *What is the work done during the expan- 
sion? 



112 ENGINEERING THERMODYNAMICS 

Work Pi (F2-F1) = 150 X 144 (3-012 - 1.506) = 32,530 

foot-pounds. 
How much heat is required? 
Heat added = H2 — {h + XiLi) 

= 11934 -.(330-2 +0.5 X 863.2) 
= 11934 — 761.8 = 431.6 B.t.u, 

PROBLEMS 

1. One pound of water is raised in temperature from 60° to 90° F. 
What is the increase in entropy? Ans. 0.056. 

2. Dry and saturated steam has a pressure of 100 lbs. per sq. in. absolute. 
{a) What is the entropy of the Uquid? Ans. 0.4743. 

(b) What is the entropy of evaporation ? Ans. 1.1277. 

(c) What is the total entropy of the steam ? Ans. 1.6020. 

3. Steam at 150 lbs. per sq. in. absolute has a quahty of 0.90. 

(a) What is the entropy of the liquid? Ans. 0.5142. 

(b) What is the entropy of evaporation? Ans. 0.9490. 

(c) What is the total entropy of the steam ? Ans. 1.4632. 

4. Steam having a temperature of 300° F. has an entropy of evapora- 
tion of 1. 1900. What is its quality? Ans. 0.994. 

5. Steam having a pressure of 200 lbs. per sq. in. absolute has a total 
entropy of 1.5400. 

{a) What is the total entropy of dry and saturated steam under the 
given pressure ? Ans. 1.5456. 

(b) Is the steam wet or dry ? 

(c) What is its quality? Ans. 0.994. 

6. Steam having a pressure of 125 lbs. per sq. in. absolute is super- 
heated 100° F. 

(a) What is its total entropy ? Ans. 1.6484. 

{b) What is the total entropy of dry and saturated steam under the 

given pressure ? Ans. 1.5839. 

(c) What is the entropy of the superheat ? Ans. 0.0645. 

7. Steam having a pressure of 150 lbs. per sq. in. absolute has a total 
entropy of 1.6043. 

(a) What is the total entropy of dry and saturated steam under the 

given pressure ? ' Ans. 1.5692. 

(b) Is the above steam saturated or superheated? How can you tell? 

(c) How much superheat has the steam ? Ans. 50° F. 

8. A boiler generates steam of 0.90 quality at a temperature of 350° F. 
with the feed water admitted at 90° F. What is the increase in entropy? 

Ans. 1.3594- 



CHAPTER VIII 



PRACTICAL STEAM EXPANSIONS AND CYCLES 



Efficiency of an Engine Using Steam Without Expansion. In 

the early history of the steam engine, nothing was known about 
the ''expansive" power of steam. Up to the time of Watt in 
all steam engines the steam was admitted at full boiler pressure 
at the beginning of every stroke and the steam at that pressure 
carried the piston forward to the end of the stroke without any 
diminution of pressure. Under these circumstances the volume 
of steam "used at each stroke at boiler pressure is equal to the 
volume swept through by the piston. 




— ^ Volume 
Fig. 34. — Indicator Diagram of an Engine using Steam without Expansion. 

An indicator diagram representing the use of steam in an 
engine without expansion is shown in Fig. 34. This diagram 
represents steam being taken into the engine cyHnder at i at 
the boiler pressure. It forces the piston out to the point 2 
when the exhaust opens and the pressure drops rapidly from 
2 to 3. On the back stroke from 3 to 4 steam is forced out of 
the cylinder into the exhaust pipe. At 4 the pressure rises 

113 



114 ENGINEERING THERMODYNAMICS 

rapidly to that at i due to the rapid admission of fresh steam 
into the cyHnder. In this case the thermal efficiency (E) is 
represented by 

J, _ work done _ (Pi - F^) (V2 - Vi) ^ .^. 

heat taken in 778 (xiLi -\- h — Ih) 

where the denominator represents the amount of heat taken in, 
with the feed water at temperature of exhaust, ^. In actual 
practice the efficiency of engines using steam without expansion 
is about 0.06 to 0.07, when the temperature of condensation is 
about 100° F. When steam is used in an engine without ex- 
pansion and also without the use of a condenser the value of 
this efficiency is still lower. It will be observed that under the 
most favorable conditions obtainable the efficiency of an engine 
without expansion cannot be made under normal conditions to 
exceed about 7 per cent. 

In the actual Newcomen steam engines efficiency was very 
much lower than any of the values given because at every stroke 
of the piston a very much larger amount of steam had to be 
taken in than that corresponding to the volume swept through 
by the piston on account of a considerable quantity of steam 
condensing on the walls of the cylinder. 

Quality of Steam During Adiabatic Expansion. A very im- 
portant equation, having wide application in steam engineering 
will now be developed. This equation is used in finding the 
quality of steam after adiabatic expansion, and can easily be 
derived after a further study of the T-cf) diagram. 

Referring to Fig. 32 (page 106), the line TiC represents the in- 
crease of entropy due to the latent heat added during the steam- 
ing process. If this steaming process had stopped at some 
point such that the steam was wet, having a quality x, this con- 
dition of the steam could be denoted by the point s, where 

TiC 

This relation is obvious, for a distance along TiC represents 
the entropy of steam, which is proportional to the latent heat 



PRACTICAL STEAM EXPANSIONS AND CYCLES 1 15 

added, which in turn is proportional to the amount of dry steam 
formed from one pound of water. In like manner, ~— is the 
quality of steam that has been formed along T2d at temperature 

It is thus apparent that any point on a 7-0 diagram will give 
full information in regard to the steam. The proportional dis- 
tances on a line drawn through the given point between the 
water and the dry steam lines and parallel to the X-axis give 
the quality of the steam as shown above. The ordinate of the 
point gives the temperature and corresponding pressure, while 
its position relative to other lines such as constant volume and 
constant total heat lines which can be drawn on the same dia- 
gram, will give further important data. Such a point (as m) 
is said to be the " state point " of the steam. 

For the present we are particularly interested in the quality of 
steam during adiabatic expansion. Since in adiabatic expansion 
no heat transfer takes place, the entropy remains constant, 
and, therefore, on a 7-0 diagram, this condition is represented by 
a straight vertical line as cgf (Fig. 32) or smk. Equation 75, 
(page 107), must be modified for wet steam as follows: 

L T L 

<i> = e + x- = loge ;^ + xi -r, (77) 

-L i ^1 

where x = the quality or dryness fraction of the steam, and 
4> = the total entropy of dry saturated steam, as before. 

Since in adiabatic expansion the entropy remains constant, 
the following equation can be written 

01 = 02 (total entropies) 
or 

ei + xi^ = e2 + x2^' (78) 

Knowing the initial conditions of steam, the quality of the 
steam at any time during adiabatic expansion can be readily 
determined. Thus, suf)pose the initial pressure of dry satu- 
rated steam to be 100 pounds per square inch absolute, and the 



Ii6 ENGINEERING THERMODYNAMICS 

final pressure after adiabatic expansion 17 pounds per square 
inch. 

From the steam-tables we find that the total entropy 0, for 
dry steam at 100 pounds pressure is 1.6020; that is^ 

4)1 = 1.6020. 

The entropies at 17 pounds pressure are also obtained from the 
tables, and we have, substituting in equation (78) 

1.6020 = 0.3229 + 0C2 1.4215 
whence, 

0C2 = 0.899. 

For a rapid and convenient means of checking the above 
result, the " Total Heat-entropy " diagram (see Marks and Davis' 
Steam Tables, Diagram I), can be used. From the intersection 
of the 100-pound pressure line and that of unit quality ("satura- 
tion fine"), is dropped a vertical line (fine of constant entropy 
= 1.602) to the 17-pound pressure line. This latter intersec- 
tion is found to lie on the 0.90 quahty line. 

Graphical Determination of Quality of Steam by Throttling 
Calorimeter and Total Heat-entropy Diagram. It will be re- 
membered that the throttling calorimeter (pages 74-77) depends 
for its action upon the fact that the total heat of steam which ex- 
pands without doing work remains the same, the heat in excess 
of that required to keep the steam dry and saturated going to 
superheat the steam. Suppose that steam enters the calorimeter 
at a pressure of 150 pounds per square inch absolute, and is throt- 
tled down to 17 pounds per square inch, the actual temperature 
being 240° F. Since the saturation temperature for steam at 
17 pounds pressure is 219.4, the steam in the calorimeter is 
superheated 240° — 219.4° or 20.6 degrees. In order to find the 
quahty of the live steam refer to the " M oilier Diagram " (Total 
Heat-entropy Diagram, Fi'g. 35) and find the intersection of the 
20.6 degrees superheat line with the 17-pound pressure Hne. 
From this point follow a horizontal line (line of constant total 
heat) to the left until it intersects the 150-pound pressure line. 
This point of intersection is found to He on the 0.96 quaKty line. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 



117 



Formula 65 (page 77) gives the following result in close 
agreement with the diagram: 

1153.1 + 0.47 (240 - 21Q.4) - 330-2 
863.2 
= 0.965. 



Xi = 




Entropy 
Fig. 35. — Mollier Diagram for Determining QuaKty of Steam. 

Construction of Adiabatic Curve by Total Heat-entropy Dia- 
gram. By means of this " Mohier " diagram the curve for 
adiabatic expansion can be very readily drawn on a pressure- 
volume diagram when the initial quahty of steam at " cut-off " 
is known. Assume that one pound of wet steam at the initial 
condition of pressure and quahty as determined above is ad- 
mitted to the engine cyHnder per stroke, and that there is previ- 
ously in the clearance space 0.2 cubic feet of steam (see Fig. 
36), at exactly the same' condition. The volume of a pound of 
dry saturated steam at this initial pressure of 150 pounds is, from 



ii8 



ENGINEERING THERMODYNAMICS 



the steam-tables, 3.012 cubic feet. At 0.965 quality it will be 
I (0.965 X 3.012) = 2.905 cubic feet. On the scale of abscissas 
this amount added to the 0.20 cubic feet in clearance gives 



• Clearance Steam 
^Admission 



I Cut-ofE 




0.20 3.15 

Volume of Steam in Cylinder (Cu. ft.) 

Fig. 36. — Illustrative Indicator Diagram of Engine Using Steam with Expansion. 



400- 



•53100 

r 

03 

a 

t 
o 



) 


/ V- 










/ 




\ 


Supeiheaied 
W Steajn 








G 


\ 


g' 


/ 


r 








\ 

s 




M 


D 


F 




f' 



0.130 



.566 



1.528 



Fig. 37. 



Entropy — (p 
•Temperature-entropy Diagram of Steam Engine. 



3.105 cubic feet, the volume to be plotted at cut-off. Other 
points in the adiabatic expansion curve can be readily plotted 
after determining the quality by the method given on page 116. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 119 

The use of the temperature-entropy diagram in exhibiting 
the behavior of steam in an adiabatic expansion and the various 
heat exchanges in the passage of steam through a steam cycle 
will now be discussed and illustrated with a practical example. 

Fig. 37 illustrates the heat process going on when feed water 
is received in the boilers of a power plant at 100° F., is heated 
and converted into steam at a temperature of 400° F., and then 
loses heat in doing work. When the feed water first enters 
the boiler its temperature must be raised from 100° to 400° F. 
before any " steaming " begins. The heat added to the Hquid 
is the area MNCD. This area represents the difference between 
the heats of the liquid (374 — 68) or about 306 B.t.u. The 
horizontal or entropy scale shows that the difference in entropy 
between water at 100° and 400° F. is about 0.437.* 

Every reader should understand how such a diagram is con- 
structed and especially how the curves are obtained. In this 
case the curve NC is constructed by plotting from the steam- 
tables the values of the entropy of the liquid for a number of 
different temperatures between 100° and 400° F. 

If, now, water at 400° F. is converted into steam at that tem- 
perature, the curve representing the change is necessarily a 
constant temperature line and therefore a horizontal, CE. Pro- 
vided the evaporation has been complete, the heat added in the 
"steaming" process is the latent heat or heat of evaporation 
of steam {L) at 400° F., which is approximately 827 B.t.u. 

The change in entropy during evaporation is, then, the heat 
units added (827) divided by the absolute temperature at which 
the change occurs (400 + 460 = 860° F. absolute) or 
r 827 

r = 865 = °-9^^- 

The total entropy of steam completely evaporated at 400° F. 
is, therefore, 0.566 -f 0.962, or 1.528.! To represent this final 

* As actually determined from Marks and Davis' Steam Tables (pages 9 and 
15), the difference in entropy is 0.5663 — 0.1295 or 0.4368. Practically it is im- 
possible to construct the scales in the figure very accurately. 

t Entropy like the total heat {H), and the heat of the liquid {h) is measured 
above the condition of freezing water (32° F.). 



I20 ENGINEERING THERMODYNAMICS 

condition of the steam, the point E is plotted where entropy 
measured on the horizontal scale is 1.528 as shown in the figure.* 
The area MNCEF represents, then, the total heat added to a 
pound of feed water at 100° F. to produce steam at 400° F., 
and then the area OBCEF represents, similarly, the total heat 
{H in the steam- tables) above 32 degrees required to form one 
pound of steam at 400° F. 

Adiabatic Expansion and Available Energy. A practical ex- 
ample as to how the temperature-entropy diagram can be used 
to show how much work can be obtained by a theoretically per- 
fect engine from the adiabatic expansion of a pound of steam 
will now be given. When steam expands adiabatically — with- 
out a gain or loss of heat by conduction — its temperature falls. 
Remembering that areas in the temperature-entropy diagram 
represent quantities of heat and that in this expansion there is 
no exchange of heat, it is obvious that the area under a curve of 
adiabatic expansion must be zero ; this condition can be satisfied 
only by a vertical line which is a line of constant entropy. 

The work done during an adiabatic process, while it cannot be 
obtained from a ''heat diagram," can very readily be determined 
from the area under the adiabatic curve of a pressure-volume 
diagram, or better still by the use of steam- tables as follows : In 
an adiabatic expansion the amount of work done is the mechani- 
cal equivalent of the loss in internal energy as explained in Chap- 
ter III. Therefore, it is only necessary to determine the internal 
energy of the steam at the beginning and end of the adiabatic 
expansion. 

IiH = h-\- ocJiL, 
I2H = h + 0C2 I2L' 

* The point E is shown located on another curve RS, which is determined 
by plotting a series of points calculated the same as E, but for different pressures. 
If more heat has been added than was required for evaporation, the area DCEF 
would have been larger and E would have fallen to the right of RS, indicating by- 
its position that the steam had been superheated. The curve RS is therefore a 
"boundary-line" between the saturated and superheated conditions. This curve 
can also be plotted from the values obtained from a table of the entropies of dry 
saturated steam. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 121 

Work during adiabatic expansion = loss in internal energy 

= Qii + xiliL — h. — 0C2I2L) 77^? in foot-pounds. 

For the case in Fig. 37 the adiabatic expansion curve will He 
along the line EF, and if the temperature falls to 100° F., the ex- 
pansion will be from E to G, and during this change some of the 
steam has been condensed. If now heat is removed from this 
mixture of steam and water until all the steam is reduced to the 
liquid state, but without further lowering of the temperature, 
the horizontal line GN * will represent the change in its condi- 
tion. The quantity of heat absorbed in this last process, techni- 
cally known as condensing the steam, is represented by the 
area MNGF. The difference between this heat rejected by the 
steam at 100 degrees and the total heat added above 100° F. is 
the '* available energy " of the steam,t and is represented by 
the area NCEG. By means of diagrams like those in the pre- 
ceding figures, it will now be shown how the available energy 
of dry saturated steam for any given conditions can be readily 
calculated from the data given in the steam- tables. 

Fig. 38a is a temperature-entropy diagram representing dry 
saturated steam which is expanded adiabatically from an ini- 
tial temperature Ti, corresponding to a pressure Pi, to a lower 
final temperature T2 corresponding to a pressure P2. The other 
initial and final conditions of total heat (H) and entropy (0) 
are represented by the same subscripts i and 2. The available 
energy or the work that can be done by a perfect engine under 

* That the steam might have been dry and saturated, the expansion would 

have had to follow the curve ES and G would have appeared at G'. The heat of 

the liquid, h, of a pound of steam at 100° F. is represented by OBNM, and the 

heat of evaporation (L) is MNG'F', so that the total heat (h-\-L or H) is OBNG'F'. 

The total heat of wet steam is expressed hy h-\- xL, where x is the quality or 

relative dryness. In the case of this adiabatic expansion, then, h is as before 

OBNM and xL is MNGF. It is obvious also that the hnes NG and NG' have 

the same relation to each other as the areas under them, so that 

lineNG area MNGF xL NG 

T- — ^if7=r/ = ---.■-■-_.-,. = —r, or .-■-■^ . = x (see pa^e 118) 

ImeNG' area MNG'F' L' NG' v fs j 

showing that the quality of the steam at any point, G, on a constant temperature 
line is given by the ratio of NG to NG'. 

t It is equal to the net work of the Rankine cycle, and depends upon the ini- 
tial and final conditions of the steam. 



122 



ENGINEERING THERMODYNAMICS 



these conditions is the area NCEG. It is now desired to obtain 
a simple equation expressing this available energy £„ in terms 







C/ Ti Pi \ 


i,E 








\ 






/ 


^^^^^^^^^ 


o\ 


p' 


B 






N 







F 




F'' 



Entropy <p^ <}>„ 

Fig. 38a. — Temperature-entropy Diagram for Dry Saturated Steam 
Expanded Adiabatically. 

of total heat, absolute temperature and entropy. Explanations 
of the preceding figures should make it clear that 

Hi = area OBNCEF, 

H2 = area OBNG'F', 

Ea = area (OBNCEF + FGG'FO - OBNG'F', 

Ea = Hi-H2 + (ct>2- 0l) r2.* • (79) 

An appHcation of this equation will be made at once to deter- 
mine the heat energy available from the adiabatic expansion 
of a pound of dry saturated steam from an initial pressure of 
165 pounds per square inch absolute to a final pressure of 15 
pounds per square inch absolute. 

Example. Pi = 165 ri = 826 degrees, from steam- tables. 
P2= 15 T2= 673.0 

Hi = 1195.0 B.t.u. 
H2 = 1150.7 B.t.u. 
' </)i = 1. 5615 

<h = 1.7549 

* It should be observed that this form is for the case where the steam is ini- 
tially dry and saturated. For the case of superheated steam a slightly different 
form is required which is given on page 125. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 123 

Substituting these values in equation (79), we have 

Ea = II95-0 - 1150-7 +(i-7549 - i-S^is) 673 = i74-5 ^-t-u. 
per pound of steam. 

Now if in a suitable piece of apparatus like a steam turbine 
nozzle, all this energy that is theoretically available could be 
changed into velocity, then we have by the well-known formula 
in mechanics, for unit mass,* 

y2 

— = Ea (foot-pounds) = Ea (B.t.u.) X 778, 

2g __ 

V = V778 X2gEa= 223.8 VEa, (80) 

where V is the velocity of the jet and g is the acceleration due to 
gravity (32.2), both in feet per second. 

Solving then for the theoretical velocity obtainable from the 
available energy we obtain the following: 



V = 223.8 V174.5 = 223.8 X 13.22 = 2956 feet per second. 

The important condition assumed as the basis for determining 
equation (79), that the steam is initially dry and saturated, must 
not be overlooked in its apph'cation. There are, therefore, two 
other cases to be considered: 

(i) When the steam is initially wet, 

(2) When the steam is initially superheated. 

Available Energy of Wet Steam. The case of initially wet 
steam is easily treated in the same way as dry and saturated 
steam. Fig. 38b is an example of the case in hand. At the ini- 
tial pressure Pi the total heat of a pound of wet steam (hi -f XiLi) 
is represented in this diagram by the area OBNCE''F''. The 
initial quality of the steam (xi) is represented by the ratio of the 

CE'' 

lines — — — . The available energy from adiabatic expansion from 
CE 

the initial temperature Ti (corresponding to the pressure Pi) to 

the final temperature T^ (corresponding to the pressure P2) is the 

* See Church's Mechanics of Engineering, page 672, or Jameson's Applied Me- 
chanics and Mechanical Engineering, vol. I, page 47. 



124 



ENGINEERING THERMODYNAMICS 



area NCE'^G''. If we call this available energy Eaw, we have by 

manipulation of the areas, 

Ea^ = area OBNCEF + FGG'F' - OBNG'F' - G"E''EG, 

E.,, = Hi-H2 + (ch-ct>i)T2-{cl>i-<}>,){T,-T2) * (8i) 

or, 

Ea^ = H^ - H2 + {cf>2 - <h) T2 - ^{i - x,){Ti - T2). (82) 





C/ Ti P, E 


"\e 


N 

/ 


/ T, P, g" 


!g \g' 


f" 


If y 



Entropy (p^ (p^ 



4>„ 



Fig. 38b. — Temperature-entropy Diagram of Wet Steam Expanded Adiabatically. 

The velocity corresponding to this energy is found by substi- 
tution in equation (80) , just as for the case when the steam was 
initially dry and saturated. 

Example. Calculations for the velocity resulting from adia- 
batic expansion for the same conditions given in the preceding 
example, except that the steam is initially 5 per cent wet, are 
given below. 

Ti = 826 degrees from tables 
T2 = 673.0 degrees " 
Hi = 1195.0 B.t.u. " 

H2 = 1150.7 " 
' 01 = 1.5615 

02 = 1.7549 

Li = 856.8 B.t.u. 
Xx = i.oo — 0.05 = 0.95 

<tl = ^ + ^1, 0x = Xl ^ + ^1, <^l — 0a: = :p (l — »l)- 
il i\ J-\ 



Pi = 165 lbs. absolute. 
P2 = 15 lbs. absolute. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 125 

Eav, = (1195-0 - 11507) + (1-7549 - 1-5615) 673 - -f^ 

X (i - 0.95) (826 - 673) = 44-3 + 130.2 - 7-93 
Eaw = 166.5 B.t.u. per pound of wet steam. 

V = 223.8 VeZ = 223.8 X V166.5 

= 223.8 X 12.9 = 2886 feet per second. 

This result can be checked very quickly by the ''total heat- 
entropy" or ''MoUier" diagram included in Marks and Davis' 
and some other steam-tables. The intersection of the 0.95 
quality line and 165 pounds pressure line is found to lie on the 
1 152 B.t.u. total heat line. Since the expansion is adiabatic, 
the entropy remains constant. Therefore, following the verti- 
cal or constant entropy Hne (entropy = about 1.507) down to 
its intersection with the 15 pounds pressure Kne, we find that 
the total heat at the end of adiabatic expansion is 985 B.t.u. and 
1152 — 985 = 167 B.t.u. available energy, as above. 

The velocity can be readily checked by the scale at the left of 
the diagram. 

If the steam were superheated to begin with, the available 
energy during adiabatic expansion could be obtained in the 
same way by means of the diagram. 

Available Energy of Superheated Steam. The amount of 
energy that becomes available in the adiabatic expansion of 
superheated steam is very easily expressed with the help of Fig. 
39. Two conditions after expansion must be considered: 

(i) When the steam in the final condition is superheated, 

(2) When the steam in the final condition is wet (or dry 
saturated) . 

Using Fig. 39 with the notation as before except E^s is the avail- 
able energy from the adiabatic expansion of steam initially super- 
heated in B.t.u. per pound, (f>s and Hs are respectively the total 
entropy and the total heat of the superheated steam at the 
initial condition, then obviously from the diagram, when the 
steam is wet at the final* condition, 

E„, = H,-H2 + (<A2 - 0.) T2. (83) 



126 



ENGINEERING THERMODYNAMICS 



When the steam is superheated at the final condition, 

Eaa = Hs — H2 — (03 — 02 ) 2^2 . (84) 

It will be observed that these equations (83) and (84) are the 
same in form as (79), and that equation (83) differs only in hav- 
ing the terms Ha and <f>s in the place of Hi and 0i. In other 



LHs 





Ti Hi 


/ 


) 




/ T^ 


\ 


i' 






/ T2 




\ 


\.H. 




*. 


*: 


i 


% 





Entropy- (^ 
Fig. 39. — Temperature-entropy Diagram for Superheated Steam. 

words equation (79) can be used for superheated steam if the 
total heat and entropy are read from the steam tables for the 
required degrees of initial superheat. 

The following examples illustrate the simplicity of calcula- 
tions with these equations: 

Example i. Steam at 150 pounds per square inch absolute 
pressure and 300° F. superheat is expanded adiabatically to i 
pound per square inch absolute pressure. How much energy in 
E.t.u. per pound is made available for doing work ? 



PRACTICAL STEAM EXPANSIONS AND CYCLES 127 

Solution. Hs = 1348.8 B.t.u. per pound, 
H2 = 1103.6 " " " 
<^ = 1.980, 

<f>s= 1.732, 

r2 = 559-6" F., 

Eas = 1348.8 - IIO3.6 + (1.980 - 1.732) 559.6 

= 383.9 B.t.u. per pound. 

The result above may be checked with the total heat-entropy 
chart in Marks and Davis' Steam Tables and Diagrams (Dia- 
gram I), and obtain thus (1349 — 967) or 382 B.t.u. per pound. 

Example 2. Data same as in preceding example except that 
the final pressure is now 35 pounds per square inch absolute. 
(Final condition of steam is superheated.) Calculate E^i. 

Solution. Hs = 1348.8 B.t.u. per pound, 
H2' = 1166.8 " " 

<t>s= 1.732, 

<^' = 1.6868, 
r/ = 718.9° F., 

Eas = 1348.8 - 1166.8 - (1.7320 - 1.6868) 718.9 
= 149.5 B.t.u. per pound. 

The Rankine Cycle.* In Chapter IV it was shown that the 
Carnot cycle gave the maximum efficiency obtainable for a heat 
engine operating between given units of temperature. In order 
that a steam engine may work on a Carnot cycle, the steam 
must be evaporated in the cylinder instead of in a separate 
boiler, and condensed in the cylinder, instead of being rejected 
to the air or to a separate condenser. Such conditions are 
obviously impracticable, and it has, therefore, been found nec- 
essary to adopt some other cycle which conforms more with 
practical conditions. The most efficient practical steam cycle, 
and the one which has, therefore, been adopted as the standard 
with which the efficiency of all steam engines may be compared, 
is the Rankine Cycle. The pressure-volume diagram of this cycle 

* Also known as the Clausius Cycle, having been published simultaneously 
and independently by Clausius. 



128 



ENGINEERING THERMODYNAMICS 



is shown in Fig. 40. Steam is admitted at constant pressure and 
temperature along ab. At b cut-off occurs, and the steam ex- 
pands adiabatically from b to c, some of it condensing during the 
process. The steam is then discharged at constant pressure 
and temperature along the back pressure line cd. Line da 
represents the rise in temperature and pressure at constant 
volume when the inlet or admission valve opens. 




Volume 

Fig. 40. — Indicator Diagram of Ideal Rankine Cycle. 



The four stages of the Rankine cycle may also be stated as 
follows: 

(i) Feed water raised from temperature of exhaust to tem^ 
perature of admission steam. (Line da.) 

(2) Evaporation at constant admission temperature. (Line 
ab.) 

(3) Adiabatic expansion down to back pressure. (Line be.) 

(4) Rejection of steam at the constant temperature corre- 
sponding to the back pressure. (Line cd.) 

Following the usual method for calculating the net work done 
in a cycle, we have, assuming one pound of steam: 

Wab = YT 8" (^i^i^i) = external work of evaporation, Ei 

(in B.t.u.), 
Wbc = loss in internal energy 

= hi + xJli - (^2 + 0C2IL2) B.t.u., (85) 

Wdc = - 7T8 (PcVc) = - yys (^2F2-'^2) = external work of 

evaporation at temperature of exhaust (B.t.u.), 
W,, = o. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 129 

Adding, 

Net work of cycle, W = jjg ^i^^i^i + ^h + xJu — h — ^^m 

but 

yy-g- PiFi + Il\ = Li (equation (62), page 66). 

Therefore, 

W = hi-\- XiLi -h- 0C2L2 (B.t.u.). (86) 

This means that the net work of the Rankine cycle is equal to 
the difference between the total heat of the steam admitted and 
the total heat of the steam exhausted. This statement applies 
whether the steam is initially wet, dry or superheated, and, there- 
fore, it becomes a very simple matter to determine the work 
done in a Rankine cycle by referring to the total heat-entropy 
diagram. 

It should be noted that this net work of the Rankine cycle is 
the so-called ** available energy " of steam as defined and dis- 
cussed in the preceding section, and that equation (82) will give 
the same result as equation (79). (The student should check 
this statement.) 

Fig. 41 is the r-(/) diagram for a Rankine cycle using dry 
saturated steam to begin with. The letters ahcd refer to the 
corresponding points in the pressure- volume diagram. The net 
work of the cycle is B -)- C, which is the difference between the 
total heats at admission and exhaust. The heat added per cycle 
is A -f B -1- C + D and the 

Thermal efficiency = ^ j^ ^ j^ q + D ^^^^ 

_ h + ociLi — Jh — OC2L2 
hi -f- XiLi + h 

It should be carefully observed that /^ in the denominator 
must always be subtracted from hi + XiLi (the total heat above 
32° F.), in order to give. the total heat above the temperature of 

* The final quality can be determined by equation (78). 



I30 



ENGINEERING THERMODYNAMICS 



feed water, which in engine tests is always assumed for the 
purpose of comparison to be the same as the exhaust temperature. 




Entropy— 
Fig. 41 . — Temperature-entropy Diagram of Rankine Cycle. 

One pound of steam at a pressure of 160 lbs. per sq. in. abso- 
lute and quality of 0.95 performs a Rankine cycle exhausting at 5 
lbs. per sq. in. absolute. 

What is the quality of the exhaust ? 

Solution. The total entropy at the initial condition 

= .5208 + .95 X 1.0431- 

The total enthropy at the exhaust 

= .2348 + 1.6084X. 

Then .5208 + .95 X 1.043 1 = -2348 + i.6o84:t:. 

From which rr = 0.794. 

What is the net work of the cycle ? 
Work = Hi-H2 = 335.6 + .95 X 858.8 - 130.1 - .794 
X 1000.3 ~ 227.2 B.t.u. 

What is the efficiency of the cycle ? 

2 27.2 



Efficiency = 



- = .222 or 22.2 per cent. 



335.6 +.95X858.8-130.1 

The Practical or Actual Steam Engine Cycle. In the steam 
engine designed for practical operation it is impossible to expand 



PRACTICAL STEAM EXPANSIONS AND CYCLES 



131 



the steam down to the back-pressure Hne; and, furthermore, it 
is evident that some mechanical clearance must be provided. 
The result is that in the indicator diagram from an actual steam 
engine, we have to deal with a clearance volume, and both incom- 
plete expansion and incomplete compression as shown in Fig. 
42. In order to calculate the theoretical efficiency of this prac- 




Volurae 
Fig. 42. — Indicator Diagram of Practical Engine Cycle. 

tical cycle, it is necessary to assume that the expansion Hne cd 
and the compression Hne fa are adiabatic. Knowing then the 
cyHnder feed of steam per stroke and the pressure and volume 
relations as determined from the indicator diagram, one can 
calculate the theoretical thermal efficiency by obtaining the net 
area of the diagram (expressed in B.t.u.) and dividing by the 
heat suppHed per cycle. In order to obtain the net area of the 
diagram, the latter may be divided up into several simple parts 
as follows: 

gcdi — a Rankine cycle, 

idej — a rectangle, 

hafj — a Rankine cycle (negative), 
gbah — a rectangle (negative) . 

The above areas can be evaluated in B.t.u. by methods pre- 
viously explained (see pages 113 and 129), and thus the net work 
of the cycle can be determined. 



132 ENGINEERING THERMODYNAMICS 

The heat added per cycle is equal to 
Total heat of cylinder feed (b to c) — cylinder feed X heat of 
Hquid of feed water at the temperature of the exhaust. 
Exercise. Calculate the thermal efficiency for the ''practi- 
cal " indicator diagram shown in Fig. 42, having given the fol- 
lowing data: 

Cylinder feed = i cu. ft., 
Initial steam pressure = 165 lbs. per sq. in. absolute (dry 
saturated), 
Exhaust steam pressure = 2 lbs. per sq. in. absolute. 

PROBLEMS 

1. What is the entropy of the liquid of steam of 92 per cent quality at a 
pressure of 15 lbs. per sq. in. absolute? Ans. 0.313. 

2. With a quality of 0.90, what is the entropy of evaporation of steam 
at a pressure of 25 lbs. per sq. in. absolute? Ans. 1.224. 

3. If there is 1.2 lbs. of water in 8 lbs. of wet steam, what is its quality? 

Ans. 0.85. 

4. What is the total entropy of steam of 94 per cent quahty at a pres- 
sure of 100 lbs. per sq. in. absolute? Ans. 1.534. 

5. It was found that an engine operating under a pressure of 155 lbs. 
per sq. in. absolute was using steam at a temperature of 561° F. What was 
the condition of the steam? Ans. Superheat 200° F. 

6. If steam at 200 lbs. per sq. in. absolute, 95 per cent quality, is caused 
to expand adiabatically to 228° F., what are the properties at the lower 
point ? (That is, final total entropy, entropy of evaporation, quality and 
volume.) Ans. 1.495; i-i6oo; 0.831; 16.7 cu. ft. 

7. One pound of steam at a pressure of 100 lbs. per sq. in. absolute 
and a quality of 50 per cent is expanded isothermally until it is dry and 
saturated. Find the heat supplied and the work done. 

Ans. 444.0 B.t.u. and 31,890 ft. -lbs. 

8. What will be the final total heat of dry saturated steam that is ex- 
panded adiabatically from 150 lbs. per sq. in. absolute down to 10 lbs. 
per sq. in. absolute? Ans. 999.7 B.t.u. 

9. Steam having a quality of 20 per cent is compressed along an adia- 
batic curve from a pressure of '20 lbs. per sq. in. absolute to a pressure whose 
temperature is 293° F. What is the final quality? Ans. 15.4 per cent. 

10. Determine the final quality of the steam and find the quantity of 
work performed by 2 lbs. of steam in expanding adiabatically from 250 lbs. 
per sq. in. absolute pressure to 100 lbs. per sq. in. absolute, the steam 
being initially dry and saturated. Ans. 93.4 per cent; 12948 B.t.u. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 133 

11. Assume that i lb. of steam of a pressure of 160 lbs. per sq. in. abso- 
lute and a quality of 95 per cent performs a Rankine cycle, being exhausted 
at a pressure of 5 lbs. per sq. in. absolute. Compute the quality of the steam 
exhausted and the efficiency of the cycle. Find the final volume. 

Ans. 79.4 per cent; 22.2 per cent; 58.2 cu. ft. 

12. What is the work of a Rankine cycle if the steam initially at 
200 lbs. per sq. in. absolute pressure, superheated 200° F., goes through such 
a cycle with a back pressure of i lb. per sq. in. absolute? If, instead of ex- 
pansion in such a cycle, the steam (for same conditions) were to expand in 
a turbine nozzle, what would be the velocity of the steam leaving ? 

Ans. 385 B.t.u.; 4390 ft. per sec. 

13. One pound of the steam at a pressure of 100 lbs. per sq. in. absolute 
with a quality of 0.90 performs a Rankine cycle exhausting at a back pres- 
sure of 2 lbs. per sq. in. absolute. 

What is the net work of the cycle? Ans. 235 B.t.u. 

What is the efficiency of the cycle? Ans. 23.4 per cent. 

14. Two pounds of steam at a pressure of 125 lbs. per sq. in. absolute 
and a volume of 8.34 cu. ft. performs a Rankine cycle. The exhaust pres- 
sure is 25 lbs. per sq. in. absolute. 

What is the net work of the cycle? Ans. 260 B.t.u. 

What is the efficiency of the cycle? Ans. 12.5 per cent. 



Application of Temperature-entropy Diagram to Analysis of 
Steam Engine. The working conditions of a steam engine, as 
stated before, can be shown not only by the indicator card, but 
also by the empIo3anent of what is known as a ''temperature- 
entropy" diagram. These diagrams represent graphically the 
amount of heat actually transformed into work, and in addition, 
the distribution of losses in the steam engine. 

For illustration, a card was taken from a Corliss steam engine 
having a cylinder volume of 1.325 cubic feet, with a clearance 
volume of 7.74 per cent, or 0.103 cubic feet; the weight of steam 
in pounds per stroke (cyHnder feed plus clearance) was 0.14664 
pounds. Barometer registered atmospheric condition as 14.5 
pounds per square inch. The scale of the indicator spring used 
in getting the card was 80 pounds to the inch. Steam, chest 
pressure was taken as 153 pounds per square inch (absolute), 
and a calorimeter determination showed the steam to be practi- 
cally dry and saturated. 



134 ENGINEERING THERMODYNAMICS 

The preliminary work in transferring the indicator card to 
a T-(j) diagram, consists first in preparing the indicator card. 
It was divided into horizontal strips at pressure intervals of lo 
pounds with the absolute zero line taken as a reference; this 
line was laid off 14.5 pounds below atmospheric conditions. 
(See Fig. 43a.) For reference, the saturation curve was drawn. 
Having known the weight of steam consumed per stroke and 
specific volume of steam (from the Steam Tables), for various 
pressures taken from the card, the corresponding actual vol- 
umes could be obtained; this operation is, merely, weight of 
steam per stroke multiplied by specific volume for some pressure 
(0.14664 X column 5 in the table below), the resulting value be- 
ing the volume in cubic feet for that condition. These pressures 
and volumes were plotted on the card and the points joined, re- 
sulting in the saturation curve, 2" -d" -%"-()" . 

The next step consisted in constructing a " transformation '* 
table with the columns headed as shown. All the condensing and 
evaporation processes are assumed to take place in the cyhnder 
and the T-<^ diagram is then worked up for a total weight of 
one pound of steam as is customary. Column i shows the re- 
spective point numbers that were noted on the card; column 2, 
the absolute pressures for such points; column 3, the corre- 
sponding temperatures for such pressures; column 4, the vol- 
ume in cubic feet up to the particular point measured from the 
reference Hne of volumes; column 5, the specific volume of a 
pound of dry and saturated steam at the particular pressure 
(Steam Tables); column 6, the volume of actual steam per 
pound, obtained by dividing the volumes in column 4 by 0.14664 
pound (total weight of steam in cyHnder per stroke); column 7, 
the dryness factor '^x," found by dividing column 6 by column 5; 

column 8 is the entropy of evaporationf— jfor particular condi- 
tions (Steam Tables) ; column 9 is the product of column 7 and 
column 8; column 10 is the entropy of the liquid at various 
conditions as found in the Steam Tables; column 11 is the sum 
of column 9 and column 10, giving the total entropy. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 



135 




















ENTROPY 


DIAGRAM 




















. 








_A 


/ 
/ 


B 

—r 


1' 


__ 


._. 




._. 


5' 
"h- 


.._. 


4^ 


! 










son 


: 








/ 
/ 


1 


P22 


23 






2 




31 

6 
s/ 


t 




1 
1 
1 


\ 










ocn 


- 






rl 


/ 
/ 


f20 














10 
IT. 


i 


^ 


1 
1 
1 

1 


\ 










^ 200 


- 






/ 
/ 
/ 
/ 


I 






^ 


^ 


^ 


< 






1 

1 

1 




1 
1 
1 

1 

1 




V 








ft — 
150^ 


- 




1 


/ 




Q^ 


G 


^^ 








L_. 


— 





1 
.1 — 

f' 


— 


1 
1 




V 


\ 

V, 


\ 
\ 




t 


I8|l7 


16 15 


14 




: 
100- 


/ 


1 

1 
1 


1 


























1 










r 






































\' 


Kn 


"r( 










































At 


L? 










































- 


- 






1 




















1 



















M 


M 




5 






' 


1 
1. 


1 



1 






1. 


5N 








2. 


J 



Entropy 
Fig. 43a. — Temperature-entropy Diagram of Actual Steam Engine Indicator 

Diagram. 



136 



ENGINEERING THERMODYNAMICS 



TRANSFORMATION TABLE 



I 


2 


3 


4 


5 


6 


7 


8 


9 


10 


II 


a 




IS 


S.I 

i 

>8 


III 


2^ S ^ 


C.2 
Q 






12 

0. 

S 
+-> 


Is 


I 


145 


356 


0.1050 


3-II 


0.716 


0.2302 


I .0612 


0.2440 


0.5107 


0.7547 


2 


140 


353 


0.2250 


3 


22 


1-536 


0.4770 


1.0675 


. 5090 


0.5072 


I. 0162 


5 


120 


341 


0.4230 


3 


73 


2.890 


0.7740 


1-0954 


0.8475 


0.4919 


1-3394 


9 


48 


279 


. 8500 


8 


84 


5.810 


0.6580 


1-2536 


0.8250 


0.4077 


1.2327 


12 


20 


228 


1.4225 


20 


08 


9.700 


0.4830 


1-3965 


0.6740 


0.3355 


1.0095 


15 


8 


183 


0.9758 


47 


27 


6.660 


0.1410 


1-5380 


0.2168 


0.2673 


0.4841 


18 


8 


183 


0.3500 


47 


27 


2.390 


0.0506 


1.5380 


0.0778 


0.2637 


0.3451 


20 


30 


250 


0.1825 


13 


74 


1-245 


0.0907 


I-33II 


0.1208 


0.3680 


0.4888 



Above table is employed for transferring the P-V diagram to the T-(f> 
diagram. 

After this table was completed, columns 3 and 1 1 were plotted. 
Convenient scales were selected, the ordinates as temperatures 
and the abscissas as entropies. The various points, properly 
designated, were connected as shown on the T-(j) diagram, the 
closed diagram resulting. This area shows the amount of heat 
actually transformed into work. This diagram is the actual 
temperature-entropy diagram for the card taken and may be 
superimposed upon the Rankine cycle diagram in order to de- 
termine the amount and distribution of heat losses. 

The water line, A-A\ and the dry steam line, C-C , were 
drawn directly by the aid of Steam Table data, i.e., the entropy 
of the Kquid and the entropy of the steam taken at various 
temperatures, and plotted accordingly. 

Before the figure could be studied to any extent, the theoreti- 
cal (Rankine) diagram had to be plotted, assuming that the steam 
reaches cut-off under steam-chest conditions; that it then ex- 
pands adiabatically down to back pressure and finally exhausts 
at constant pressure to the end of the stroke without compres- 
sion. This diagram is marked, A-C-E-H, on the T-(j> plane. 
The steam-chest pressure of 153 pounds per square inch absolute 



PRACTICAL STEAM EXPANSIONS AND CYCLES 137 

fixes the point, C, when the temperature line cuts the steam line, 
C-C. The rest of the cycle is self-evident. 

Referring to the T-^ diagram, Fig. 43a, H-A-C-E is the Ran- 
kine operation with no clearance for one pound of working fluid. 
The amount of heat supplied is shown by the area, Mi-H-A-C-N, 
and of this quantity, the area Mi-H-E-N * would be lost in 
the exhaust while the remainder, H-A-C-E, would go into 
work. This is theoretical, but in practice there are losses, and 
for that reason, the Rankine cycle is used merely for comparison 
with the actual card as taken from a test. The enclosed irregu- 
lar area, 1-2-3 • • • 22-23, is the amount of heat going into 
actual work. By observation, it is evident that a big area re- 
mains; this must represent losses of some sort or other. That 
quantity of work represented by the area, 1-5-5'-!', is lost on 
account of wire-drawing; the area $'-C-D-F, shows a loss due 
to initial condensation; the loss due to early release is shown 
by the area F-12-14-F' for the real card, and by D-G-E for the 
modified Rankine cycle (such a loss, in other words, is due to 
incomplete expansion); that quantity represented by 2 2-^-1 '-i 
is lost on account of incomplete compression, and H-A-B-iS is 
the loss due to clearance. The expansion line from 5 on to 9 in- 
dicates that there is a loss of heat to the cylinder walls, causing 
a loss of entropy; from 9 on to 10, re-evaporation is taking place 
(showing a gain of entropy). 

All of the heat losses are not necessarily due to the transfer 
of heat to or from the steam, as there may be some loss of steam 
due to leakage. In general, however, the T-cf) diagram is satis- 
factory in showing heat losses. 

Fig. 43a was constructed for the purpose of showing how the 
actual thermal efficiency and the theoretical thermal efficiency 
(based on the Rankine cycle) can be obtained from the T-<l) 
diagram. The letters in Fig. 43a refer to the same points as 
in Fig. 43b, the only difference between the two diagrams 

* The areas MiHACN and MiHEN should have added to them, the rectangu- 
lar area between Mi — N, and the absolute zero line (Fig. 216), which is not shown 
on the diagram (Fig. 43a). 



138 



ENGINEERING THERMODYNAMICS 



800 



700 



500 

a' 




^ 400 



Ml 



Fahr. Line 



300 



200 



100 



Thermal Eff . = 



_ W 



M'iHaJcn' 



liiLLM 



N 



1.6 



2.0 



.4 .8 1.2 

Entropy 

Fig. 43b. — Temperature (absolute) -entropy Diagram of Actual Steam Engine 

Indicator Diagram. 



.PRACTICAL STEAM EXPANSIONS AND CYCLES 139 

being the addition of the absolute zero temperature line to 
Fig. 43b. 

Work done 



Thermal efficiency = 
Actual thermal efficiency = 



Heat added 
Shaded area W 



Mi'H-A'C-N' 

^ ^5 — 1 — _i_ (pianimeter) 
26.00 sq. m. 

= 0.096, or 9.6 per cent.* 

Rankine cycle efficiency (theoretical thermal efficiency) 

E-A-C-E 

Mi'-H-A-C-N'' 

= ^f^ - (Planimeter) 
26.00 

= 0.203, or 20.3 per cent. 

Discussion of Heat Engine Efficiencies. The action of all 
heat engines is, in general, to produce motion against a resistance 
or to perform work, that is to say, engines or machines by whose 
agency one form of energy is converted into an energy of another 
kind. The heat engine may be a motive power engine, in which 
energy in the form of heat is converted into energy in the form 
of mechanical work; or it may be one in which mechanical 
energy is transformed into the potential energy of a gas. 

All these engines depend for their action on the variation in 
volume, temperature and pressure of some gas or vapor, or upon 
the explosion of some gaseous mixture. The gas or vapor which 
undergoes these changes in a heat engine is called the working 
fluid. In an air compressor, the working fluid is air. In the gas 
engine, work is done by the expansion of the products of com- 

* This value may be regarded as the actual thermal efficiency for one stroke, 

inasmuch as in the succeeding strokes the "cushion" steam will be used over and 

over again and hence will not constitute a heat loss. The actual thermal efficiency 

W 
of a steam engine under runnirtg conditions would then be given by „, _ p^,w 

M 18 nCiv 

where 18 BCE is a Rankine cycle with clearance and complete compression. 



I40 ENGINEERING THERMODYNAMICS 

ibustion of a mixture of air and combustible gas. In the oil en- 
gine, the action is the same except that oil is used. The steam 
engine, the engine with which we are mostly concerned, uses the 
T-apor of water for its working fluid. 

In all these heat engines the working fluid undergoes a regular 
series of changes, the same change occurring in the same order, 
over and over again. So, we can say, the working fluid goes 
through a circuit or ** cycle " of operation. Beginning at a par- 
ticular condition it passes through a series of successive states 
of pressure, volume and temperature and returns to the initial 
condition. Thus, the engine has started from a certain point, 
has gone through a regular series of movements in a fixed order, 
and has returned to the place from which it started. Such a 
series of regular changes is called a cycle. 

The simplest form of reversible cycle is the Camot wherein 
the working fluid undergoes four reversible processes between 
the temperatures Ti and T2 (page 45) . 

All heat engines work between some range of temperature, as 
in this case, Ti to T2, and in view of this fact, the available 
relation of heat is, as explained in Chapter III, 

Ex=^^^ (89) 

The heat represented by the temperature range between Ti 
and T2 is the only heat which under any consideration can be 
turned into work. It represents the ** maximum possible effi- 
ciency " (El), In fact, an ideal steam engine is impossible. 
Such an engine with this cycle cannot be actually constructed 
because of the various thermodynamic losses. 

The above expression, however, represents the efficiency of 
the Carnot cycle and merely signifies that it is the highest value 
that is obtainable for such a cycle working between such tem- 
perature limits. 

The amount of work performed by the working fluid upon the 
piston of an engine is obviously equal to the mechanical equiva- 
lent of the difference between the heat supplied (Hi) and the 



PRACTICAL STEAM EXPANSIONS AND CYCLES 141 

heat abstracted (^^2); and the theoretical thermal efficiency 
(E2), of the cycle is therefore given by the expression, 

E. = ^^- (90) 

We know that Ei is the highest possible efficiency that is 
attainable. £2, the theoretical thermal efficiency of the cycle, 
must then approach Ei as its Kmit. So the ratio of the one to 
the other gives what is known as the " type efficiency " (E3). 

E,=f. . (91) 

The closer the type efficiency approaches unity, the better 
the cycle should be. 

It is a known fact that steam engines have very large heat 
losses, — losses that cannot be entirely removed. If for a cer- 
tain amount of work (W) actually accomplished (equivalent 
heat units), the amount of heat used in the production of that 
work is Ha, then the " actual thermal efficiency " (E4) is stated 
as 

E.= ^. (9.) 

Theory implies that there is an efficiency that the actual 
thermal efficiency must approach as a limit. This relation be- 
tween the actual thermal efficiency (£4) and the theoretical 
thermal efficiency (£2) is known as the practical efficiency (E5) 
and shows how nearly the theoretical efficiency is approached, by 
the actual efficiency, or 

E. = |. (93) 

Most important of all efficiencies is that for the " mechanical 
efficiency," meaning the comparison of the useful work per- 
formed with the amount hi work theoretically possible to obtain 
with a perfect machine. In other words, in a heat engine the 



142 



ENGINEERING THERMODYNAMICS 



mechanical efficiency (Ee), is the ratio of the brake horse 
power to the indicated horse power, or 

b.h.p. 
i.h.p. 



Efi 



(94) 



Another efficiency often applied is what is known as the 
"overall efficiency" (E7), and is the relation of the output, or 
the useful work performed, divided by the heat supplied. Such 
an efficiency can be expressed in various ways, as 

E7 = El X E3 X E5 X Ee, 
= E2 X E5 X Ee, 
= E4 X Ee. 

It is noted that the "overall efficiency" is the product of 
several efficiencies, and in view of that fact, the increasing of 
any of the other six respective efficiencies means a relative in- 
crease in the ''overall efficiency." 





a 


h 




2 


\ 






3 


i 






^ 


K 










\d 


^""^^--^ c 




1 
i 


1 





a h 


S 


a'; 








1 




* 




1 






S 


1 









J 




1 


p- 


I 




\ 


a 


Id 












H 


1 1 
/ 1 

/ 1 


\ 



m 



P 



Volume 

Fig. 44a. — Pressure- volume Diagram of 
Carnot Cycle of Steam Engine. 



o m n 

Entropy— 
Fig. 44b. — Temperature-entropy 
Diagram of Carnot Cycle of 
Steam Engine. 



Referring to the Carnot cycle for a steam engine (Fig. 44a), 
the amount of work (W) actually done is shown by the area 
abed on the pressure- volume diagram and also on the tempera- 
ture-entropy diagram (Fig. 44b). Of course the area mabcn (Fig. 
44a) represents the maximum work possible with the heat (^"1) 
that was supplied, while madcn represents the amount of heat 
{H-^ that was abstracted from the cycle. This abstraction of 



PRACTICAL STEAM EXPANSIONS AND CYCLES 143 

heat is a loss, so that the net work done is the difference be- 
tween mabcn and mdcn, or area abed. The same is expressed 
by symbols of heat units as Hi — H2. 

Taking the Carnot cycle on the temperature-entropy dia- 
gram (Fig. 44b), the efficiency (Ei) is 

„ abed , . , , Hi — n2 

El = — ; — , which equals — — 

raabn Hi 

If in this cycle heat was not added along the admission line 
ab but along some other line such as a'b, the heat added would 
be decreased by the triangular area a'ab, and consequently the 
work done would be correspondingly reduced. Hence in this 
case, the efficiency of the modified cycle (Ei') is 

abed — a'ab 



El' 



mabn — a'ab 



Such a deerease in the heat supplied means an actual de- 
erease in effieiency. From this analysis, one can say that for 
the obtaining of the maximum efficiency (Ei), heat must either 
be added or subtracted along a constant temperature line, such 
as ab and not a'b, and Ti and T2 which govern the range of 
temperatures should be separated as widely as possible. This 
statement m_ust not be interpreted, however, to mean that when 
heat is added with a varying temperature (finally, attaining a 
higher value), that it may not be more efficient than one in 
which heat is added at a constant temperature of lower value. 
Cycles of such kinds are numerous. From an inspection of 
Fig. 45 it can be seen that abed is more efficient than mnop. 
Hence the greatest efficiency is obtained when all the heat is 
added at the highest possible temperature and all the heat re- 
moved at the lowest possible. This is true for all cycles. 

Thus in Fig. 46, with saturated steam, the efficiency of the 
steam engine having such a temperature-entropy diagram, would 

be ; and with superheated steam, a slight increase in 

fdabg 

efficiency is observed, as expressed by , . The relations, 

f dabb h 



144 



ENGINEERING THERMODYNAMICS 



— — — and — — - , are practically equal. Now, if the little area 
fdabg fdaeh j -i i 

heW is added to both numerator and denominator of the last ex- 
pression the efficiency for the superheated cycle is found. This 
adding of the triangle beb' has the same relative effect as the 




Entropy • 
Fig. 45. — Temperature-entropy 
Diagram Showing Heat Added 
at Increasing Temperature. 




Eatropy-0 
Fig. 46. — Temperature-entropy Dia- 
gram for Engine Using Superheated 
Steam. 



subtracting of the triangle a'ab in the Camot cycle (Fig. 44b). 
In the latter case the efficiency was increased while in the former 
it was diminished. Hence we can say that superheated steam 
will increase the theoretical thermal efficiency of the engine. 

Various explanatory solutions for efficiencies of thermal ma- 
chinery, such as steam engines, steam turbines, " locomobiles " 
units, gas producers and engines, blast-furnace gas engines, oil 
engines and pumping engines will now be given. 

I. Simple Non-condensing Steam Engine. An engine devel- 
oping 135 i.h.p. and using dry saturated steam at 125 pounds 
per square inch absolute, running with a back pressure of 0.2 
pound per square inch gage and with a barometer pressure 
equal to 14.8 pounds per square inch, uses 32 pounds of steam 
per i.h.p. per hour. At end of expansion, the pressure is 20 
pounds per square inch gage. The brake horse power is 125. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 145 

Find the various efficiencies (£1, £2, £3, £4, ^5, Ee and £7) and 
also the number of thermal units consumed per i.h.p. per 
minute. 

(a) Maximum Possible Efficiency (Ei). 

El = — ^-^; — -, where Ti is the absolute temperature of the 

steam at the initial pressure condition, and T2 is the absolute 
temperature of the steam at atmospheric. 

(344.4 + 460) - (213 + 460) 

^1 = . — 7 

344.4 + 460 

= ^ = 0.1014 or 16.14 per cent. 
804.4 



(b) Theoretical Thermal Efficiency (E2). 
£2 = 



Hi — H2 



Hi 

_ (hi + XiLi) -{h + X2L2) + IM (P4 - P5) X :^F4 ^ 
(hi + XiLi) - h 

The quantity (h + iCiLi), the total heat, is read from the 
"Mollier Diagram,"* the steam being at an initial condition 
of 125 pounds per square inch 
absolute with a quality Xi of 
unity. From this point the ex- 
pansion takes place adiabatically 
to the end of expansion at 34.8 
pounds per square inch absolute, 
where the quality X2 is found to volume 

be 0.9225 and the total heat ^^^- ^^' ^'^^^^^^ ^'"^^ ^""^^ 
(^2 + oc^L^ is 1093 B.t.u.; P4 is 

the pressure at end of expansion, P5 is absolute back pressure, 
F4 is the volume corresponding to the pressure P4 (from Steam 
Tables) and fe, the heat of the liquid at the back pressure 
(14.8 + 0.2) or 15 pounds per square inch absolute, then 

* Diagram I, Marks and Davis' Steam Tables and Diagrams. 




146 ENGINEERING THERMODYNAMICS 

P> ^ 1 189 - 1093 + yH X (29.8 - 20) X 0.9225 X 11.89 
^ I189 - 181 

_ 1 189 - 1093 + 19-85 
I189 - 181 ' 

115-85 

= — "^—f- = O.I 15 or 1 1.5 per cent. 
1008. 

(c) Type Efficiency (E3). 

= , = 0.713 or 71.3 per cent. 
16.14 

(d) Actual Thermal Efficiency (£4). 

where (W) is the amount of work actually accomplished per hour 
and (Ha) is the amount of heat used in developing that work, then 

£, = ms^ , 

32(1189-181) 

^545 0.0788 or 7.88 per cent. 



32 X 1008 
(e) Practical Efficiency (£5). 
£4 



^-S' 



= -^ — = 0.686 or 68.6 per cent. 
II. 5 

(f) Mechanical Efficiency (Eq). 

J. b.h.p. 

l.h.p. 

= — - = 0.926 or 92.6 per cent. 
135 

* Both numerator and denominator will be expressed in terms of B.t.u. per hour. 
The numerator is the heat equivalent of one horse-power-hour or (33,000 X 60) -j- 
778 = 2545 B.t.u. per hour. The denominator is the net heat used at the rate of 
32 pounds of steam per hour. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 147 

(g) OveraU Efficiency (£7). 

Et = E4 X Eq 

= 0.0788 X 0.926 

= 0.073 or 7.3 per cent. 

For a check, by using another formula, we have 

Et = E2XE^X Ee 

= 0.115 X 0.686 X 0.926 = 0.073 or 7.3 per cent. 

(h) Heat units per i.h.p. per minute. 

We know that one i.h.p. represents 33,000 foot-pounds of 
work, and furthermore that one B.t.u. equals 778 foot-pounds. 
From this relation we find that the number of B.t.u. per i.h.p. 

per mmute (no losses) is^^^^^ — - — , or 42.42. 

778 

But the actual thermal efficiency (£4) is but 7.88 per cent, so 

the actual number of B.t.u. necessary per minute will equal 

t^,' °^ 538 B.tu. 

0.0705 

II. Compound High-speed Non-condensing Steam Engine. 

Assume that this engine works under the same pressure condi- 
tions as the above engine with the steam quality at unity. 
Find all efficiencies with the number of B.t.u. required per i.h.p. 
per minute when it operates at 130 i.h.p. and uses 25 pounds of 
steam per hour. Prony brake test gives no b.h.p. 

Note. The values for the efficiencies, Ei, Ei and E3 will be the same for this 
engine as for the case of the simple engine as just calculated. 

(a) Maximum Possible Efficiency (£1) = 16.14 per cent. 

(b) Theoretical Thermal Efficiency (£2) = n-S per cent. 

(c) Type Efficiency (£3) = 71.3 per cent. 

(d) Actual Thermal Efficiency (E4). 

£, = , ^545 * ^ 

25 (1189 — 181) ^ 

= "^-^ — - = o.ioi or 10. 1 per cent. 

25^ X 1008 ^ 

* See page 146 (foot-note). 



148 ENGINEERING THERMODYNAMICS 

(e) Mechanical Efficiency (Eg). 

i.h.p. 
= x¥7 = 0-847 or 84.7 per cent. 

(f ) Practical Efficiency (£5) . 

^' = ^ 

= — — = 0.878 or 87.8 per cent. 

' o 

(g) Overall Efficiency (£7). 

E7 = E^XE6 

= o.ioi X 0.847 = 0.0854 or 8.54 per cent. 

(h) Heat per i.h.p. per minute. 

Refer to the former problem for the method of obtaining the 
value of 42.42 B.t.u. per i.h.p. per minute (no losses). 

B.t.u. per i.h.p. per minute = ? 

O.IOI 

= 420 B.t.u. 

III. Steam Tixrbine. A turbine using steam at 200 pounds 
per square inch absolute pressure at 160° F. superheat and a 
back pressure of 0.5 pound per square inch absolute with a 
barometer of 29.92 inches (of mercury), consumes 12 pounds of 
steam per kilowatt-hour. Find maximum possible efficiency, 
the theoretical thermal efficiency, the type efficiency, actual 
thermal efficiency and the practical efficiency, for such condi- 
tions. Calculate the heat consumption per kilowatt-minute and 
also per electrical horse-power-minute. 

(a) Maximum Possible Efficiency (Ei) . 



Ei = 



^ (541 >9 + 460) - (80 4- 460) 
541.9 -h 460 

= -^ — — = 0.461 or 46.1 per cent. 
1001.9 



PRACTICAL STEAM EXPANSIONS AND CYCLES 149 

(b) Theoretical Thermal Efficiency (£2), 

1288 - 886 

1288 - 80 
where 1288 is the total heat in B.t.u. per pound at 200 pounds per 
square inch absolute, from which point the expansion takes place 
adiabatically down to 0.5 pound per square inch absolute. The 
total heat at the lower pressure is 886 B.t.u. per pound. The 
heat of the Hquid at 0.5 pound per square inch absolute 
(back pressure) is 80 B.t.u. per pound, or, 

^ = "^^ = 0-333 or 33.3 per cent. 
1200 

(c) T3^e Efficiency, £3 = tt * 

-Ss = '"^^ ■ = 0.722 or 72.2 per cent. 
0.461 

W 

(d) Actual Thermal Efficiency, E^ = -—- 

Then, E ^-^^ 



12 (1288 -80) X 0.746 



Note. Denominator is multiplied by the value 0.746 because the number of 
horse power multiplied by this coefficient gives the equivalent power in kilowatts. 

P - 2545 

12 X 1208 X 0.746 
= 0.226 or 22.6 per cent. 

E± 

(e) Practical Efficiency, £5 = — * 

E2 

(f) Heat per kilowatt-minute, 

£5 = — ^ = 0.679 or 67.9 per cent. 
33-3 

We know that there are 42.42 B.t.u. per horse power per 
minute (no losses) and that one horse power equals 0.746 kilo- 
watt. Then, similarly for no losses, we have 

\. = '^6.6 B.t.u. per kilowatt per minute. 
0.746 ^ ^ ^ ^ 

* See page 146 (foot-note). 



I50 ENGINEERING THERMODYNAMICS 

But the test shows an actual thermal efficiency of 22.6 per 
cent and the actual heat unit consumption must be established 
on this basis, thus, 

B.t.u. per kilowatt per minute = -^-~ = 2';o B.t.u. 
^ ^ 0.226 ^ 

The heat per electrical horse power per minute can readily be 
found by multiplying the B.t.u. per kilowatt-minute by 0.746 
(which is the kilowatt equivalent of a horse power), or 250 X 
0.746 = 187 B.t.u. per electrical horse power per minute. 

IV. Unit Consisting of Steam Boiler, Steam Engine and 
Electric Generator. A '^ locomobile " engine * gives the follow- 
ing test results: i.h.p. = 185; kilowatts of generator =115; 
steam consumption per i.h.p. per hour = 10 pounds; steam pres- 
sure = 210.3 pounds per square inch gage; superheat = 260° F.; 
barometer, 14.7 pounds per square inch; vacuum = 12.7 pounds 
per square inch (nearly 26 inches of mercury) ; feed water tem- 
perature = 142° F. 

Find the various efficiencies, Ei, E2, £3, E^, E5 and Eq and de- 
termine the number of B.t.u. per i.h.p. per minute. The coal 
that was used in the boiler was found to have 14,350 B.t.u. per 
pound. With an evaporating capacity of 9 pounds of steam per 
pound of coal, what is also the boiler efficiency? 
(a) Maximum Possible Efficiency (£1), 

Ti - T2 (Ti + ^i) - T2 
'" Ti " T^ + h ' 
where h is the degrees Fahrenheit of superheat; f other values 
are the same as in previous problems, then, 

J. (391.9 H- 260 -{- 460) - (126 -f- 460) 

391.9 -f- 260 + 460 
^ _ 1111.9- 586 
mi. 9 
525-9 



0.473, or 47.3 per cent. 
I III. 9 

* A "locomobile" is a combination of steam boiler and engine, arranged with 
the engine placed on top of the boiler. 

t The 126° F. corresponds to (14.7 — 12.7), or 2 lbs. per sq. in. pressure. (See 
Steam Tables.) 



PRACTICAL STEAM EXPANSIONS AND CYCLES 15 1 

(b) Theoretical Thermal Efficiency (£2), 

j^ Hi — H2 

^^ 1325-1005 ^ 
^3^5 - 94 

where 1325 B.t.u. is the heat read from the "MoUier Diagram " 
at 225 pounds per square inch absolute and 260° F. superheat. 
From this point the steam expands adiabatically to the lower 
condition of 2 pounds per square inch absolute (14.7 — 12.7) 
where the total heat is 1005 B.t.u. per pound; the value 94 is 
the heat of the liquid at 2 pounds per square inch absolute 
pressure. Then, 

T^ ^20 . , 

IL2 = — ■ = 0.26, or 26 per cent. 
1231 

(c) Type Efficiency (£3). 

£3 = • = 0.55, or 55 per cent. 

47-3 

(d) Actual Thermal Efficiency (£4). 



E,= 



2545 



10 X (1325 - 94) 



= ^ = 0.2666, or 20.66 per cent. 

10 X 1231 

(e) Practical Efficiency (E5). 

j^ 20.66 , 

£5 == — — = o-794j or 79.4 per cent. 
26.00 

(f) Mechanical Efficiency (Eq). 

^ b.h.p. kilowatt output 
i.h.p. indicated kilowatts ' 

where ^'kilowatt output " of the power developed by the gener- 
ator and the " indicated kilowatts " are the number of kilowatts 
corresponding to i.h.p. Then, 

^' = 185 X 0.746 = °-^33' "' ^3-3 per cent. 



152 ENGINEERING THERMODYNAMICS 

(g) B.t.u. per i.h.p. per minute. 

B.t.u. per i.h.p. per minute = = 2ot:. 

^ ^ 0.2066 ^ 

(h) Efficiency of the Boiler. 

r^rr Lbs, steam evap. per lb. coal X B.t.u. avail, per lb. of steam.* 

B.t.u. per lb. of coal 

^ 9 [1325 - (142 -32)] 
14,350 

o X 1215 ^ ^ 

= ~ == 0.7625, or 76.25 per cent. 

14,350 

V. Gas-producer and Engine. If the mechanical efficiency 
of a producer-gas engine is known to be 85 per cent, what is the 
i.h.p., provided the machine is running under Prony brake test 
at 658 r.p.m. with a load of 400 pounds (net) at an 8-foot radius? 
Suppose the producer has an efficiency of 60 per cent, find the 
B.t.u. per i.h.p. per minute and also the B.t.u. per b.h.p. per 
minute if the coal consumed per i.h.p. per hour is i pound. 
Heating value of the coal is 14,000 B.t.u. per pound. 

(a) Calculation of Indicated Horse Power. 

Mechanical Efficiency (Ee) = .' ' ■ , or i.h.p. = '"' > 

I.h.p. Ae 

, , 2 X3141 XrXN XW 

D.n.p. = J 

33,000 

which is the equation for b.h.p., where r is the radius of the brake 
in feet, N is the number of revolutions per minute of the engine, 
and W is the net load on the brake arm at the measured radius, 
then, 

= 2X3.141X8X658X400 ^ ^^^. 
33,000 



also, 



. , 400 

* B.t.u. per pound above the temperature of feed water. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 153 

(b) B.t.u. per i.h.p. per minute. 

This value is dependent on the actual thermal efficiency (£4), 
thus, since 

E,=- 2545 ., 

0.00 X 1. 00 X 1400 
= 0.303, or 30.3 per cent. 

.*. B.t.u. per i.h.p. per minute = — '■ — = 140. 

0-303 

(c) B.t.u. per b.h.p. per minute. 

Since the mechanical efficiency is 85 per cent, B.t.u. per b.h.p. 

per minute = —^ , or 164.8. 
0.85 

VI. Blast-furnace Gas Engine. The results of a test of a 
blast-furnace gas engine show that 100 cubic feet of gas is con- 
sumed per i.h.p. per hour. Heating value of gas was 120 B.t.u. 
per cubic foot; the i.h.p. developed 800 and b.h.p. = 580. Find 
the mechanical efficiency {E^ of the machine. Find the actual 
thermal efficiency (£4). What are the heat equivalents B.t.u. 
per i.h.p. per minute, and also per b.h.p.? 

(a) Mechanical Efficiency {E^. 

Eq = ^ = 0.725, or 72.5 per cent. 
000 



(b) Actual Thermal Efficiency (E4). 

£4 = '^^^ = 
120 X 100 

(c) B.t.u. per i.h.p. per minute. 



£4 = ■ = 0.212, or 21.2 per cent. 

120 X 100 



B.t.u. per i.h.p. per minute = = 200. 

0.212 

(d) B.t.u. per b.h.p. per minute. 

Since the mechanical efficiency was found to be 72.5 per cent, 

B.t.u. per b.h.p. per minute = , or 276. 

0.725 

Vn. Oil Engine. In tests of an oil engine, the switchboard 
readings showed a generator output of 375 kilowatts, and the in- 



154 ENGINEERING THERMODYNAMICS 

dicator diagrams 560 i.h.p. If the machine consumed 0.32 pounds 
of oil per i.h.p. per hour, what are the efficiencies, £4 and E^} 
How many B.t.u. were required per b.h.p. and i.h.p. per minute? 
Heat value of the oil was 19,000 B.t.u. per pound. 

(a) Mechanical Efhciency {E^. 

The output of the machine is 375 kilowatts which is equal to 

■ ^'^ , or 502 '^electrical " horse power. From this, we can say, 
0.746 

C02 
Ee = ^ = 0.8975, or 89.75 per cent. 
560 

(b) Actual Thermal Efficiency (E4). 

E4 = ^ = 0.418, or 41.8 per cent. 

0.32 X 19,000 

(c) B.t.u. required per i.h.p. per minute. 

-r» • 1 • 42.42 

B.t.u. per I.h.p. per mmute = — ^ = 101.4. 

0.418 

(d) B.t.u. required per b.h.p. per minute. 

B.t.u. per b.h.p. per minute = = 113.1. 

0.8975 

Vin. Pumping Engine. A pumping engine uses 12 pounds 
of steam per i.h.p. per hour under the following conditions: 
steam pressure = 190 pounds per square inch absolute; quality 
of steam = 0.98; barometer = 14.8 pounds per square inch; 
pressure at end of expansion = 6 pounds per square inch; back 
pressure = 2.2 pounds per square inch absolute. What are the 
various efficiencies if the i.h.p. is 850 and the '' delivery " horse 
power 825? Find the number of B.t.u. per i.h.p. per minute, 
and also the " duty " per million B.t.u. 

(a) Maximum Possible Efficiency {Ei). 

^ ^ (377-6 + 460) - (130 + 460) 

377.6 + 460 

247.6 
= ^ = 0.2952, or 29.52 per cent. 

o / 



PRACTICAL STEAM EXPANSIONS AND CYCLES 1 55 

(b) Theoretical Thermal Efficiency (£2). 
Using the ''Mollier Diagram," 

1181 - 946 + 4M X (6 - 2.2) X 0.811 X 61.80 

JtL2 = 

1181 — 97 

_ 1181 - 946 + 35-3 

1181 — 97 

270.^ 
= • ' = 0.249, or 24.9 per cent. 

(c) Mechanical Efficiency (Ee). 

E^ = -—^ = 971, or 97.1 per cent. 
850 



(d) Type Efficiency (£3). 



212 
Ez = — '— = 0.851, or 85.1 per cent. 
24.9 



(e) Actual Thermal Efficiency (E^). 
E. '-^ 



12 X (1181 - 97) 
= 0.1959, or 19.59 per cent. 

(f) Practical Efficiency (£5). 

£5 = = 0.786, or 78.6 per cent. 

24.9 

(g) B.t.u. per i.h.p. per minute. 

B.t.u, per i.h.p. per minute = - = 216.4. 

0.1959 

(h) Duty per Million B.t.u. 

Pumping engines are usually reported as being capable of a 

certain *' duty," which is the amount of useful work done in 

foot-pounds per million B.t.u. supplied. Duty per milHon B.t.u. 

= 778 X 1,000,000 X Et, where £7 is the overall efficiency. 

Then, 

£7 = £4 X Eq, and in this case 

= 0.1959 X 0.971 = 0.19, or 19 per cent. 

Duty of the pump pei" million B.t.u. = 778 X 1,000,000 X 
0.19, or 147,900,000 foot-pounds. 



156 ENGINEERING THERMODYNAMICS 



PROBLEMS 

1. Steam at a pressure of 100 lbs. per sq. in. absolute having a quality 
of 0.50 expands adiabatically to 15 lbs. per sq. in. absolute. What is the 
quality at the end of the expansion? Ans. 0.50. 

2. One pound of steam at a pressure of 100 lbs. per sq. in. absolute has 
a volume of 4 cu. ft. and expands adiabaticaUy to 15 lbs. per sq. in. absolute. 

What is the quaHty at the initial and final conditions ? 

Ans. 0.905; 0.817. 
What is the work done during the expansion? Ans. 114 B.t.u. 

3. Two pounds of steam having a temperature of 330° F. and quaHty 
of 0.90 expand adiabatically to 230° F. 

What is the quality at end of expansion? Ans. 0,826. 

What is the work of the expansion? Ans. 193 B.t.u. 

4. One pound of steam having a pressure of 125 lbs. per sq. in. abso- 
lute and volume of 4.17 cu. ft. expands adiabatically to 25 lbs. per sq. in. 
absolute. 

What is the quality at the initial and final conditions ? 

Ans. 100 degrees Sup.; 0.953. 
What is the work of expansion? Ans. 95.6 B.t.u. 

5. Given the steam as stated in problem 4 but expansion complete at 
100 lbs. per sq. in. absolute. What is the quality at 100 lbs. pressure ? 

Ans. 70' F. Sup. 

6. What would the pressure be if the steam in problem 4 were expanded 
adiabatically until it became dry and saturated ? 

Ans. 56,6 lbs. per sq. in. absolute. 

7. Two pounds of steam having a pressure of 100 lbs. per sq. in. abso- 
lute and a temperature of 377.8° F. expand adiabatically to 15 lbs. per sq. 
in. absolute. 

What is the quality at the initial and final conditions? 

Ans. 50 degrees F. Sup.; 0.917. 
What is the work of the expansion? Ans. 241 B.t.u. 

8. One pound of dry and saturated steam has a pressure of 100 lbs. per 
sq. in. absolute, and expands to 20 lbs. per sq. in. absolute along a.n n = i 
curve. 

What is the quality at end of the expansion? 

Ans. 55° F. Sup. 
What is the work of the expansion? Ans. 102,750 ft.-lbs. 

What heat is required? Ans. 127.6 B.t.u. 

9. Two pounds of saturated steam at a temperature of 300° F. have a 
total volume of 12 cu. ft., and expand to a pressure of 15 lbs. per sq. in. 
absolute along an w = i curve. 



PRACTICAL STEAM EXPANSIONS AND CYCLES 



157 



What is the quahty at the initial and final conditions ? 

Ans. 0.928; 12° F. Sup. 
What is the work of the expansion ? Ans. 173,000 ft. -lbs. 

What heat is required? Ans. 309 B.t.u. 

10. One pound of steam at a temperature of 360° F. has a quaUty of 
0.50, and expands under constant pressure to a volume of 3.4 cu. ft. 
What is the quality at the final condition ? 

Ans. 90 degrees F. Sup„ 
What is the work of the expansion? Ans. 9750 ft.-lbs. 

What heat is required? Ans. 482.1 B.t.u. 

II. Two pounds of steam at a pressure of 100 lbs. per sq. in. absolute 
have a volume of 4 cu. ft., and expand under constant temperature to a 
volume of 8 cu. ft. 

What is the quality at the initial and final conditions ? 

Ans. 0.452; 0.904. 
What is the work of the expansion ? Ans. 57,600 ft.-lbs. 

How much heat is required? Ans. 401 B.t.u. 



1^5.61* 





145.61^ , ^ 

/' — ' ! 1 




j/ 1 H.P. I 


^^ 


■^I [ C.E.I 


43.36* ,___+.— ^1 "1 I 


1 i 


^TaT'Atm.iLine ' ! ! " ! 


i I ]39.36^{ 


sv.iFl 11! ! : ! 


! . 1 1 



Zero Line 



27. 96 1*' 



Atm. Line j^^"^ 


y 


C.E.I « 


.9.76* .^.^.i— — ^-'^ I ! 




■ J' 


^"H — J ! i ! 


\J\ 


i6.76#i ; j ! i ; 




!5.96*; 



Zero Line 

Fig. 48. — High and Low Pressure Indicator Diagram of Compound Steam 

Engine. 



Combined Indicator Card of Compound Engine. The method 
of constructing indicator diagrams to a common scale of vol- 
ume and pressure shows where the losses peculiar to a compound 
steam engine occur, and, to the same scale, the relative work areas. 

As the first step divide the length of the original indicator dia- 
grams into any number 'of equal parts (Fig. 48), erecting per- 
pendiculars at the points of division. 



158 ENGINEERING THERMODYNAMICS 

In constructing a combined card, select a scale of absolute 
pressure for the ordinates and a scale of volumes in cubic feet 
for the abscissas. To the scale adopted draw in the atmos- 
pheric pressure (" atm.") line, see Fig. 49. 

Lay off the low-pressure clearance volume on the ii:-axis to 
the scale selected. In like manner lay off the piston displace- 
ment of the low-pressure cylinder and divide this length into the 
same number of equal parts as the original indicator diagrams 
were divided. From the original low-pressure card (Fig. 48), de- 
termine the pressures at the points of intersection of the perpen- 
diculars erected above the line of zero pressure, taking care 
that the proper indicator-spring scale is used. Lay off these 
pressures along the ordinates (Fig. 49), connect the points and 
the result will be the low-pressure diagram transferred to the 
new volume and pressure scales. 

The high-pressure diagram is transferred to the new volume 
and pressure scale by exactly the same means as described for 
the low-pressure diagram. 

The saturation curve is next drawn. This curve represents 
the curve of expansion which would be obtained if all the steam 
in the cylinder was dry and saturated. It is very probable 
that these curves for each cylinder will not be continuous since 
the weight of the cushion steam in the low-pressure cylinder is 
usually not the same as that in the high-pressure cylinder. The 
saturation curve would be continuous for the two cards only 
when the weight of cushion steam in the high-pressure and low- 
pressure cylinders is the same (assuming no leakage or other 
losses) . 

On the assumption that the steam caught in the clearance 
spaces at the beginning of compression is dry and saturated, the 
weight of the cushion steam can be calculated from the pressure 
at the beginning or end of compression, the corresponding cylin- 
der volumes and the specific volumes corresponding to the pres- 
sure (as obtained from steam-tables). 

The total weight of steam in the cylinder is the weight of steam 
taken into the engine per stroke plus the weight of steam caught 



PRACTICAL STEAM EXPANSIONS AND CYCLES 



159 



155 


;i?^ 


...P. 


Qua 


lity ( 


^urve 100 


i 


























s 


\ 1 


^ 






i..' 
































175% 






\. 


t5.6 


f 
















































































































































































































































. 






































iH 


.P.S 


atur 


dtioi 


iCu 


rve 








































































































a 


& \c 












































! 








































































1 






j 




































\ 


































. 7 


0.61^ 


\ 


\ 






































\ 


\ 


































i 


LP.C 


\ 

^ard 


\ 
\ 




























,1( 


oi 




/ 


LV.MlE.P 
=49.5 


\' 


\ 


!l. 


P.Qi 


lalit.^ 


Cu] 


•ve 















■ 




.i 










\ 


\ 
\ 



























1 .'i 


„* 




ii 


















\ 


% 






























] 


^ 








Is: 


36^ 






























^~- 


■ 






J37. 


11*^ 






































1 
1 
































27. 

r 


6^ 






^. 


36* 




^^, 


"--.-^ 


L.P 


Sati 


irati 


one 


urve 
















L.P 


Cai 


d 




\ 


s^ 




■-> 


^-^. 


k 






















Av. 


VI.E 


P.= 


11.22 




X 


\ 


^ 


i 




Lin 


r-^- 
















L 


.76* 























__,^ 




"~"- 


—- 


~9k 


# 






\ 


^^ 































T^ 


? 













































.5 1 1.5 2 2.5 3 3.5 i 4.5 

» Volume- Cu. Ft. 

Fig. 49. — Combined Indicator Diagrams for Compound Engine. 



l6o ENGINEERING THERMODYNAMICS 

in the clearance space (cushion steam). The saturation curves 
may now be drawn by plotting the volumes which the total 
weight of steam will occupy at different pressures, assuming it 
to be dry throughout the stroke. 

The quality curve (Fig. 49) shows the condition of the steam 
as the expansion goes on. At any given absolute pressure, the 
volume up to the expansion line shows the volume of the wet 
vapor, while the volume up to the saturation curve shows the 
volume that the weight of the wet vapor would have if it were 
dry. Thus at any given absolute pressure, the ratio of the vol- 
ume of the wet vapor (as given by the expansion line of the 
indicator card) to the total volume of the dry vapor (as obtained 
from the saturation curve) is the measure of the quality of the 
steam. 

Showing the quality by the use of the figure, we have Vol. 
A —B -r- Vol. A — C = quality. By laying off this ratio from 
a horizontal Hne to any scale desired as shown, the quality curve 
may be constructed. 

Hirn's Analysis. By the use of an indicator diagram, it is 
possible to determine exactly the heat added to or lost from the 

steam. Hirn's analysis shows 
on a theoretical basis these 
gains and losses for each condi- 
tion of the steam. Investiga- 
tions have shown, however, 
that the method is of no great 
value except to designers. 
Fig. 50 is an indicator card, 

Fig. "Jo. — Indicator Diagram to Illustrate _ • i. • • . 

Him's Analysis. compression beginning at c. 

If the quality of steam at this 
point were known, the weight of and the heat in the steam can 
obviously be calculated. 

Assuming the correct quality at c is known, the heat added to 
or lost from the steam from c to d and for each point from c 
to d can be exactly determined. Laying off, for each point, 
those quantities above the zero line (Fig. 51), the shape of the 




PRACTICAL STEAM EXPANSIONS AND CYCLES 



l6l 



area cdd' is found which represents the amount of heat added 
to the steam. 

From d to a steam is taken in from the boiler. Determin- 
ing this weight from the boiler or from the condenser, the 
total weight as well as the volume at a is known, thus the en- 
tropy at a can be calculated. By comparing the conditions of 









b" 






C' A 


d' 


a ^- 


— — ^b' 


\L 


d' 

A 


c 


d 
d" 


/ 


h 


c 


d 



Areas above line represent heat added to steam 
" below " " " taken from steam 

Fig. si.— Diagram of Him's Analysis. 



the steam when entering and at a, the loss to the steam due to 
the heat stored up in the cylinder walls is shown by the area 
dd''a'a. The shape of this part of the diagram is unknown 
although its area is known if the original assumption at c is 
correct. A reasonable guess of the shape of the area may be 
made. The greater part of the initial condensation probably 
takes place at the beginning of admission and falls off very fast 
as the point of cut-off is approached, thus dd" would be very 
large and a'a very small. 

From a to b the cylinder walls give up heat to the steam which 
in turn loses its heat because of the work done. The condi- 
tion of the steam at every point along ab can be determined, and 
thus the shape of the area is also determined. 

The exhaust opens at b, the cylinder walls give up their heat 
to the steam at a greater rate than before because the steam 
is being discharged. 

The area dd^'a'a is the quantity of heat exchanged between 
the cylinder walls and the steam during the time steam is taken 
into the cylinder and the results practically mean that during 
admission so much heat must be stored up somewhere. The heat 
interchanged during expansion is worked out on the basis that 



l62 ENGINEERING THERMODYNAMICS 

if the weight and pressure of a given weight of steam are known 
at the beginning and end of a certain period, the difference in 
heat is accounted for by the work done and the heat added to 
or lost from the cylinder walls. 




Throat or smallest 
section of nozzle 



CHAPTER IX 
FLOW OF FLUIDS 

Flow through a Nozzle or Orifice. Thermodynamic prob- 
lems embrace the measurement of the flow of air or of a mix- 
ture of a liquid and vapor through a nozzle or orifice. In the 
nozzle shown in Fig. 52, let A and B be two sections through 
which the substance passes. At A let 
a pressure of P\ be maintained and at 
B a pressure of P2. To maintain the 
constant pressure at A of Pi let more 
substance be added, while at B allow 
enough of the substance to be dis- 
charged so that the constant pressure Fig. 52. —Typical Nozzle for 
of P2 is maintained. The quantity of Expanding Gases and Vapors. 

energy and the mass passing into the section A must be accounted 
for at section B and the relation of these quantities will determine 
the change of the velocity of the substance. 

After uniform conditions have been estabhshed in the nozzle, 
the same mass entering at A must be discharged at B during 
the same time. Thus any mass may be considered as a work- 
ing basis, but as a rule one pound of the substance is used. All 
formulas refer to one pound, unless another mass is definitely 
stated. 

The same quantity of energy discharged at B must enter at 
A unless heat is added to or taken from the substance between 
the sections A and B. Thus the general formula is derived: 

Energy carried by substance at B = energy carried by sub- 
stance at A + energy added between the sections A and B. 

The energy carried by the substance at the entering or dis- 
charge end of the nozzle is made up of three quantities: (i) the 
amoimt of work necessary to maintain constant pressure at each 

163 



1 64 ENGINEERING THERMODYNAMICS 

end of the nozzle; (2) the internal energy of the substance, 
(3) the kinetic energy stored in the substance because of the 
velocity which it has when passing the section. 

The amount of work necessary to maintain a constant pres- 
sure (pounds per square foot) of Pi at A or of P2 at B is -^ or 

778 

—^, where Vi and v^ are the volumes of one pound of the sub- 

778 

stance at A and B respectively. 

The internal or " intrinsic " energy in B.t.u. per pound {Ih) 
of the substance passing A or B, calculating from 32° F. is for 

A- • -1 j?x(r-4Q2)* . . 

Air or similar gases, ^ , (95) 

778 X 0.40 

Liquid, h, (96) 

Liquid and Vapor, h -\- xlL -\ (97) 

^ 2^X778/ 

Superheated Vapor, h -\-Il — -j 

Pfaup-^>sat) 
778 



+Cp(rs„p-rsat) - "^^T./"^^ , (98) 



72 
where is the kinetic energy in B.t.u. per pound of the 

2^X778 

substance as it passes a section, V = velocity in feet per second 
and g — 32.2 feet. 

If Q represents in B.t.u. per pound the heat units added to the 
substance between the sections A and B (Fig. 52), the energy 
equation for air and similar gases can be found by equating the 
total heat energy put in at A plus the energy added between 
A and B to the energy discharged at B. This general formula 
reduces to 



Fo^ - F'2 



g[^(Pi2;i-P2%) + 778e] 



2 — yi 

4 

or = 2 g X 778 [Cp (Ji - T2) + Ql (99) 

* See equation (44) , page 38. 



FLOW OF FLUIDS 165 

This thermodynamic equation is the usual form for the flow 
of air or similar gases. 

The energy equation for superheated steam can be derived 
by the use of the same general formula stated above, on the 
assumption that the substance is superheated at A and B. This 
formula reduces to 
y 2 — y 2 

— = [hi + Li + Cp (Tsnp — Tsat)l] 

2^X775 

-[h2+L2 + Cp(rgup - rsat)2] + Q, (100) 

or F2^ — Fi^ = 2 g X 778 [Tabular heat contenh^ — Tabu- 

lar heat content2 + Q]. (loi) 

From a thermodynamic standpoint, the relation between the 
initial and final condition is that of adiabatic expansion when all 
the heat which disappears as such is used in changing the veloc- 
ity, providing the nozzle is properly shaped and Q is zero. The 
diagram in Fig. 53 represents this condition of affairs on a tem- 
perature-entropy diagram for air and similar substances. The 
area acdf is Cp (T2 — 492), and the area abef is Cp (Ti — 492). 
The quantity of heat energy changed into kinetic energy is there- 
fore the area bcde and is the difference between the internal 
energy in the substance at the beginning and end of the opera- 
tion, together with the excess of work done to maintain the pres- 
sure of Pi at A over the pressure of P2 at B. The line cd would 
incline to the right if heat were added in the nozzle, since the 
effect would be to increase the velocity or increase the area cdeb. 
The line cd would incline to the left from c if heat were lost 
since the area cdeb would decrease. 

In Fig. 54 the diagram represents the conditions for super- 
heated vapor. The area aa'cdf represents the heat required to 
raise the substance from a Hquid at 32° F. to superheated vapor 
at the temperature of Tisup- The area ab'bdf is the heat con- 
tent at b, the final condition. The area a'cbb' represents, there- 

* Tabular heat content means the total heat of superheated steam as read 
from tables of the properties of superheated steam as, for example, Marks and Davis' 
Tables, pages 22-65. 



i66 



ENGINEERING THERMODYNAMICS 



fore, the heat available for increasing the velocity. The areas 
representing the heat available for increasing the velocity in Fig. 




Fig. 53. — Temperature-entropy 
Diagram of Heat Available in 
Air. 



TiSup. 




1 




Ti sat. 


a' 


/ 




T2SUP. / \ 


b 


T2sat. / \ 




/"' 




\ 


u 


T.-Ent.-Diagram 




\ 


f 






d 



Fig. 54. — Temperature-entropy Diagram 
of 'the Heat Available in Superheated 
Steam for Increasing Velocity. 



53 and Fig. 54 are shown by the cross-hatched area in Fig. 55 
and are really the representation of the work done (theoretically) 

in an engine giving such an in- 
dicator diagram. 

Evidently, the greater the 
drop in pressure, the greater 
will be the cross-hatched area 
in Fig. 55 and the greater will 
be the velocity, regardless of 
the substance. The line ab in 
Fig. 56 represents the velocity 




Fig. 



55. — Heat (Work) Available 
Increasing Velocity. 



curve with V as ordinates and — ^ as abscissas, but since with 

Pi 



FLOW OF FLUIDS 



167 



any substance expanding the weight of a cubic foot decreases as 
the pressure drops, the line cd will represent to some scale not 
here determined the weight of a cubic foot at any discharge 
pressure. 

Since the product of the area (square feet) through which the 
discharge takes place, the velocity in feet per second at the area 
and the weight (pounds) per cubic foot of the substance is equiv- 



a 
\ 






















\ 


\ 








e 














\ 




/ 


w 


;^:in.p 


-^ 


\, 












\ 


r 








\ 


\ 










/ 


\ 


,K. 








\ 




d 




/ 


/ 




^ 






^ 


^ ^ 


\ 






/ 










< 






\ 




j 


/ 




.f> 












\ 




1 




i^A 


V 










N 






/ 


/ 


/ 


















1/ 


















\ 


h 



.1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 

Fig. 56. — Illustrative Curves of Weight Discharge and Velodty. 



alent to the weight in pounds of the substance discharged per 
second. The product of the ordinates at any point of the curves 
ab and cd is proportional to the weight discharged from a pres- 
sure of Pi to a condition where the pressure is P2. The line ceb 
represents this product. Evidently there is some low pressure 
into which the weight discharged per square foot will be a maxi- 
mum, and this will be the pressure corresponding to the high 
point e on the curve. 

Weight per Cubic Foot., From the formula for adiabatic ex- 
pansion the weight per cubic foot can be obtained if the sub- 



1 68 ENGINEERING THERMODYNAMICS 

stance is similar to air.* The general formula (applied to air) is 
Piz^ii-^ = P22^2'-^ (102) 

which can be reduced to 

I P2^XPl I Pi0.286p 0.714 

- = — I ' or -=: — , (103) 

Pi'-' X RTi 

which is the weight in pounds per cubic foot of discharge. Pres- 
sures are, of course, in pounds per square foot. If the supply to 
the nozzle is from a large reservoir so that Vi can be taken as 
zero then the discharge velocity is 

V2 = V 2g X ^ (PiVi - P2V2) 
' 0.4 

F. = 109.6 v/ri[i-(^J''^]- (104) 

All quantities on the right-hand side of this equation must be 
obtained from the data of tests. Weight in pounds discharged 
through the area A (in square feet) is 



p 0.286 p 0.714 / r /p \0. 286-1 

^ = ^x RT, X 109.6 v/ri[i-(^) ]• (105) 

Maximum Discharge. This weight is a maximum when 

-— ■ = o or when P2 = 0.525 Pi. The maximum quantity of air 
dP2 

will be discharged when the low pressure is 52.5 per cent of the 

high pressure. 

Shape of Nozzle. See page 183, on the flow of steam. 

Flow of Air through an Orifice. Air under comparatively high 

* The exponent in the formula is the ratio of the specific heats (of air in this 
case). 



FLOW OF FLUIDS 169 

pressure is usually measured in practice by means of pressure 
and temperature observations made on the two sides of a sharp- 
edged orifice in a diaphragm. The method requires the use of 
two pressure gages on opposite sides of the orifice and a ther- 
mometer for obtaining the temperature ti at the initial or higher 
pressure pi. The flow of air w, in pounds per second, may then 
be calculated by Fliegner's formulas: 

w = 0.530 X f X a -^zr when pi is greater than 2 p2, (106) 

VTi 

w = 1.060 X f X a y^^-—^ — — when pi is less than 2 p2, (107) 
▼ Ti 

where a is the area of the orifice in square inches, f is a coeffi- 
cient, Ti is the absolute initial temperature in degrees Fahren- 
heit at the absolute pressure pi in the " reservoir or high-pressure 
side " and p2 is the absolute discharge pressure, both in pounds 
per square inch. When the discharge from the orifice is directly 
into the atmosphere, p2 is obviously barometric pressure. 

Westcott's and Weisbach's experiments show that the values of f 
are about 0.925 for equation (106) and about 0.63 for equation 
(107). 

For small pressures it is often desirable to substitute manom- 
eters for pressure gages. One leg of a U-tube manometer can be 
connected to the high-pressure side of the orifice and the other leg 
to the low-pressure side. Many engineers insert valves or cocks 
between the manometer and the pipe in which the pressure is to 
be observed for the purpose of '^ dampening" oscillations. This 
practice is not to be recommended as there is always the possi- 
bility that the pressure is being throttled.* A better method is 
to use a U-tube made with a restricted area at the bend between 
the two legs. This will reduce oscillations and not affect the 
accuracy of the observations. 

Discharge from compressors and the air supply for gas engines 
are frequently obtained by orifice methods. 

* Report of Power Test Committee, Journal A.S.M.E., Nov., 1912, page 1695. 



lyo ENGINEERING THERMODYNAMICS 

When pi — p2 is small compared with pi, the simple law of 
discharge * of fluids can be used as follows: 

fa 



w = — V2 g X 144 (Pi - P2) s, (108) 

144 

where f is a coefficient from experiments, g is the acceleration 
due to gravity (32.2), and s is the unit weight of the gas meas- 
ured, in pounds per cubic foot, for the average of the initial and 
final conditions of temperature and pressure. If the difference 
in pressure is measured in inches of water h with a manometer 
then 

144 (pi — P2) = — ^ X h (expressed in terms of pounds per square 
12 

foot), 

w = — 1/2 ghs X — ^ (pounds per second), (109) 

where 62.4 is the weight of a cubic foot of water (density) at 
usual " room ^^ temperatures. 

This equation can also be transformed so that a table of the 
weight of air is not needed, since by elementary thermodynamics 
144 pv = 53.3 T, where v is the volume in cubic feet of one 



If the density is fairly constant, 



144 Pl , Vl^ 144 P2 , Vo^ 

— -j — -\ , 

S 2g S 2g 

where Vi is the velocity in feet per second in the "approach" to the orifice, and Vo is 
the velocity in the orifice itself. Since Vi should be very small compared with Vo, 

Vo^ ^ 144 (pi - P2) 
2g S ' 



2 g X 144 (pl - P2) 



Vo=\/ 

w = favos = fas ^ ^ g X i44(Pi - P2 ) , 

or w = fa V2 g X 144 (pl — P2) s. 

Professor A. H. Westcott has computed from accurate experiments that the 
value of the coefi&cient f in these equations is approximately 0.60. 



FLOW OF FLUIDS 171 

pound and T is the absolute temperature in Fahrenheit. Since 
V is the reciprocal of s, then 

s = 144 p-^ 53-3 T, 
and w = 0.209 fa \/-^' (no) 

Here p and T should be the values obtained by averaging the 
initial and final pressures and temperatures. Great care should 
be exercised in obtaining correct temperatures. When equa- 
tion (47) is used, for accurate work, corrections of s for humidity 
must be made.* 

For measurements made with orifices with a well-rounded 
entrance and a smooth bore so that there is practically no con- 
traction of the jet the coefficient f in equations (47) and (48) 
may be taken as 0.98. In the rounding portion of the entrance 
to such a nozzle the largest diameter must be at least twice the 
diameter of the smallest section. For circular orifices with 
sharp corners Professor Dalby f in reporting very recent experi- 
ments stated that the coefficient for his sharp-edged orifices in a 
thin plate of various sizes from i inch to 5 inches in diameter 
was in all cases approximately 0.60; and these data agree very 
well with those pubhshed by Durley.f 

When p2 -^ pi = 0.99 the values obtained with this coefficient 
are in error less than | per cent; and when p2 -^ pi = 0.93 the 
error is less than 2 per cent. 

Receiver Method of Measuring Air. None of the preceding 
methods are adaptable for measuring the volume of air at high 
pressures as in the case of measuring the discharge in tests of 
air compressors. Pumping air into a suitably strong receiver 
is a method often used. The compressor is made to pump 

* Tables of the weight of air are given on page 181 and tables of humidity on 
page 368 in Moyer's Power Plant Testing (2d Edition). 

t Engineering (London), Sept. 9, 19 10, page 380, and Ashcroft in Proc. Insti- 
tution of Civil Engineers, vol. 173, page 289. 

% Transactions American Society of Mechanical Engineers, vol. 27 (1905), 
page 193. 



172 ENGINEERING THERMODYNAMICS 

against any desired pressure which is kept constant by a regu- 
lating valve on the discharge pipe: 

Pi and P2 = absolute initial and final pressures for the test, 
pounds per square inch. 

Ti and T2 = mean absolute initial and final temperatures, de- 
grees Fahrenheit. 

Wi and W2 = initial and final weight of air in the receiver, 
pounds. 

V = volume of receiver, cubic feet. 

PiV = WRTi, and P2V = W2RT2, where R is the constant 
53.3 for air, then weight of air pumped 

In accurate laboratory tests the humidity of the air entering 
the compressor should be measured in order to reduce this 
weight of air to the corresponding equivalent volume at atmos- 
pheric pressure and temperature. 

Principal errors in this method are due to difficulty in measur- 
ing the average temperature in the receiver. Whenever practi- 
cable the final pressure should be maintained in the receiver at 
the end of the test until the final temperature is fairly constant. 

PROBLEMS 

1. Air at a temperature of 100° F. and pressure of 100 lbs. per sq. in. ab- 
solute flows through a nozzle against a back pressure of 20 lbs. per sq. in. 
absolute. Assuming the initial velocity to be zero, what will be the ve- 
locity of discharge? Ans. 1570 ft. per sec. 

2. If the area at the mouth of the above nozzle is 0.0025 sq. ft. and the 
coefficient of discharge is unity, how many pounds of air will be discharged 
per minute? Ans. 59 pounds. 

3. What will be the theoretical kinetic energy per minute of the above 
jet assuming no frictional losses? Ans. 2901 B.t.u. 

Flow of Steam. In Fig. 52 suppose the sections A and B 
are so proportioned that the velocity of the substance passing 
section A is the same as that at section B. Such a condition 
might arise in a calorimeter or in the expansion of ammonia 



FLOW OF FLUIDS 



173 



through a throttling or expansion valve (Fig. 20), as in an ice 

machine. The pressure at A will be Pi which is greater than the 

pressure at B of P2. Fig. 57 represents the entropy diagram 

for such a condition. As the pressure falls from Pi to P2 the 

maximum heat available to produce 

velocity through the nozzle is the area 

acde. The value of the quaHty, 

represented by the symbol x for the 

substance after leaving the nozzle 

corresponds to that of point c and the 

area acde is the excess kinetic energy 

represented by the increased velocity. 

This excess kinetic energy is destroyed 

by coming into contact with the more 

slowly moving particles at B and with 

the sides of the vessel. The area acde 

is equal to {ha + XaL^ — {he + Xc Lc) 

and the relation of Xa to Xc is obviously adiabatic. The area 

ohdbf equals area oheag; thus the heat content at b is the 

same as at a. The location of b can be found as follows: 





/e p^ a 








Id 


c 


h 


/ 


I P2 






/ 

h 


T-Ent.-Diagram 









g 




f 



Fig. 57. — Diagram for no 
Velocity Change. 



Xb 



ha + XaLa — h 



(112) 



The curve shown in Fig. 58 represents the discharge of a mixture 
of steam and water {x = 0.6 at 100 pounds per square inch abso- 
lute pressure) into a vessel having the pressures shown. The 
points on this curve cannot be determined by entropy tables. 
At 100 pounds per square inch pressure the total heat of the wet 
steam is h -\- 0.6 L or 298.5 + 0.6 X 887.'6 = 831. i B.t.u. per 
pound. 

At 60 pounds per square inch absolute pressure the total heat 
may be found by the use of the entropy diagram shown in 
Fig. 59. The entropy values are taken directly from steam- 
tables. The entropy for the initial point is, then, 



ab = 0.4748 + 0.6 X 1. 1273 = 1.1512, 



174 ENGINEERING THERMODYNAMICS 

The distance 

de = ab — 0.4279 = 0.7233, 

and X at 60 pounds 

de 0.72SS 

= = ■ = 0.595. 

1. 2154 1. 2154 

2.00 



1.75 



o 

d 

o 

:S 1.25 

o 



1.00 



.75 



» .50 



^ 



.25 









■^^ 










/ 


\ 








to/ 

7 










y 












/ 










\ 


/ 










\ 


/ 












' 













10 20 30 40 50 60 70 80 90 100 
Discharge Pressure in lbs. per sq. in. 

Fig. 58. — Discharge of Steam Under Various Pressures. 

The total heat at 60 pounds pressure is 

h + 0.595 L = 262.4 + 0.595 X 914.3 = 805.4. 
The velocity of flow is 
V2 X 32.2 X 778 (83 1. 1 — 805.4) = 1135 feet per second. 
The volume of one pound is 

0.016 (i.o — 0.595) + 7-i66 X 0.595 = 4-27 cubic feet, 
and the weight per cubic foot is 



4.27 



= 0.2342 pound. 



FLOW OF FLUIDS 

The weight discharged per square inch per second is 

"35 X 0.2342 ^ ^ g ^^^^^ 
144 



175 





e h\ 



T.- Ent.- Diagram 



Fig. 59. — Temperature-entropy 
Diagram for Calculating the 
Weight of Discharge of Steam. 



Fig. 60. — Diagram Illustrating 
Radiation Loss in Nozzle. 



Velocity of Flow as Affected by Radiation. Fig. 60 shows the 
radiation losses. The condition at entrance k represented at a 
and the area acde represents the quantity of heat lost by radi- 
ation. Area aefg represents the velocity change while the point e 
represents the condition of the moving substance. 

If, after passing through the nozzle, the velocity is reduced 
to that of entrance, a point located as at b will represent the 
condition of the substance. This point would be so located that 



eb 



area aefg 



(113) 



Friction Loss in a Nozzle. Fig. 61 shows the friction loss. 
The energy converted into heat by friction varies with the square 
of the velocity. In this figure, a is the initial condition and 
aefg is the energy available for change in velocity providing 
there is no friction loss.^ The ratio of the areas acde to aefg is 
the proportional loss by friction. The point c represents the 



176 



ENGINEERING THERMODYNAMICS 



condition of the substance at the return of the friction heat to 
the substance. The heat is returned in exactly the same way as 
if it came from an outside source. The distance ch is the area 
ac(ie.-^ Lc. The area edfg represents the energy expended in 
the velocity change and the point h represents the state of the 
substance at the point of discharge. 

This condition is found existing in the fixed nozzle of most 
turbines. Point a represents the condition on the high-pressure 
side of the nozzle and point h, the low-pressure side. The abso- 
lute velocity of discharge is really caused by the energy repre- 
sented by the area edfg. 

Impulse Nozzles. Suppose that the substance is discharged 
with an absolute velocity corresponding to the area edfg (Fig. 




T.- Ent,- Diagram 




n I d 



h 



T.- Ent.- Diagram 



Fig. 61. — Diagram Illustrating 
Friction Loss in Nozzle. 



Fig. 62. ^Diagram of Heat Losses 
in a Steam Nozzle (Impulse). 



61), and that it passes into a moving nozzle, having the same 
pressure on the discharge as on the intake side. The energy 
represented by the area edfg would be used up in the following 
ways: (i) by the friction in moving nozzle; (2) residual abso- 
lute velocity; (3) and in driving the moving nozzle against the 
resistance. 

Fig. 62 shows the quantities used up by each. Point a rep- 



FLOW OF FLUIDS 177 

resents the condition of the substance before passing into the 
fixed nozzle while point h shows its condition leaving the fixed 
nozzle, the velocity corresponding to the area edfg. Area klde 
represents the energy used up in friction in the moving nozzle; 
area klnm residual velocity after leaving moving nozzle and 
area mnfg represents useful work used in moving the nozzle 
against its resistance. The condition of the substance leaving 
the nozzle is shown at q and not at h, the distance h — q being 
the area edlk divided by Lh. The substance leaves the moving 
nozzle with a velocity corresponding to the area klnm and it 
will have done work corresponding to the area mnfg. 



e V 
Fig. 63. — Impulse Nozzle and Velocity Diagrams. 

Turbine Losses. Fig. 63 is a simple velocity diagram show- 
ing, for an impulse nozzle such as occurs in many turbines, the 
relative value of those various losses. A is a stationary nozzle 
discharging against the movable blades B. The path of the steam 
is shown by the dotted line. The line db marked v represents 
the velocity of discharge of the stationary nozzle, which makes 
an angle a with the direction of motion of the moving blades. 
Call ezj the velocity of the moving blades, then h is the amount 
and direction of the relative velocity of the steam over the 
surface of the moving blades. It loses a portion of this velocity 
as it passes over the surface of the blades and lh becomes the 
actual relative velocity of discharge. The direction of lh is de- 
termined by the discharge edge of the moving blades, the angles 



178 



ENGINEERING THERMODYNAMICS 



a and jS being as shown. The residual absolute velocity is re- 
presented by r. 

The total energy equivalent of the velocity developed 

c,2 



in B.t.u. per pound = 



2g 



The residual energy per pound = 



^g 



(114) 
(115) 



Reaction Nozzles. When the substance leaving the station- 
ary nozzle passes into a moving nozzle having the pressure at 
the intake greater than at the discharge, the conditions differ 
from those just discussed. The velocity in this case is thermo- 
dynamically changed in passing through the moving nozzle. 
In the equations given in the previous discussion it was assumed 
that the moving nozzles were entirely filled with the substance, 

and when partly filled in the expand- 
ing portion, coefficients of correction 
were appHed, but in this case the 
nozzles should be so designed that 
the substance entirely fills them, as 
the corrections are unknown. 

In Fig. 64 the lines of Fig. 62 are 
reproduced together with those relat- 
ing directly to the reaction nozzle. 
Point a corresponds to the condition 
on entering the stationary nozzle, 
point h the condition on leaving it 
with a velocity corresponding to the 
Diagramof Heat Losses area cdfg. Point k represents the 
pressure at the discharge end of the 
moving nozzle, and if no friction losses or impact loss occur 
in moving nozzle, point k would represent the condition of the 
discharged substance and the area egmkhd would be accounted 
for as useful work done and residual velocity. But since friction 
losses and impact loss do occur a portion of this area edhkno can 
be set aside to represent these losses, a portion noqp represents 



a, 




1 q\ o\ e 


a 


f 1 ! 


h 


/ 1 1 ^ 




w./ ' ' 


I 


7 p n 


k 


. T.- Ent.- Diagram 





Fig. 64 
in a Steam Nozzle (Reaction) 



FLOW OF FLUIDS 1 79 

the residual velocity, while the remaining area pqgm represents 

the useful work done. 

The condition of the substance leaving the moving nozzle is 

, , ,, ,. , 1,1- area edhkno . 
given by k, the distance kl being . Area onpq rep- 

resents the residual velocity. 

Coefficient of Flow. Few experiments have been carried on 
for determining the flow of steam in nozzles proportional for 
maximum discharge or through nozzles exactly designed for an 
exact pressure. For nozzles having well rounded entrance and 
parallel portion of least diameter from 0.25 to 1.5 times the length 
of the converging entrance the coefficient of discharge is about 
1.05. For properly shaped entrances and for areas of orifices 
between 0.125 square inch and 0.75 square inch the coefficient of 
discharge varies from 0.94, the two pressures being nearly alike, 
to unity, the ratio of the pressures being 0.57. For an orifice 
through a thin plate the coefficient is about 0.82, the ratio of the 
pressures being 0.57. 

Injectors. In an injector, steam enters at A in Fig. 65 at 
the pressure of the supply. The quantity of water entering at 
C, the cross-section of the pipe, and 
the pressure of the water determine 
the pressure at ^. At Z) the pres- 
sure should be zero (atmospheric) Fig. 65.— Essential Parts of an 
or equal to the pressure in the water Injector, 

supply pipe to which D may be connected. The total hydrauKc 
head should exist as velocity head at this point. At E the pres- 
sure should be sufficient to raise the check valve into the boiler 
and the velocity sufficient to carry the intended supply into the 
discharge pipe. 

The shape of the nozzle from ^ to ^ should be such as to con- 
vert the energy in the steam at A into velocity d.t B. At B the 
water and steam meet, condensing the steam, heating the water 
and giving to the water a velocity sufficient to carry it through 
the nozzle B-D. 

All the energy accounted for at A and C must be accounted for 




l8o ENGINEERING THERMODYNAMICS 

at E. The heat lost by radiation may be neglected. The veloc- 
ity at any section of the nozzle equals 

volume passing in cubic feet 
area of section in square feet 

or F^= — , where a = area at any section corresponding to 

velocity V and a a = area at section A . 

Weight of Feed Water per Pound of Steam. Assuming the 
steam supply to be dry and reckoning from 32° F. the heat units 
contained in the steam and feed water per pound and the heat 
in the mixture of steam and feed water per pound may be easily 
calculated. 

Knowing the rise of temperature of the water passing through 
the injector and neglecting radiation losses, the pounds of feed 
water suppHed per pound of steam used by the injector may be 
obtained. Thus 

Heat units lost by steam = Kinetic energy of jet + Heat 
units gained by feed water. 

The term expression ''kinetic energy of jet " may be neglected 
since it is very small, then, 

H -hf = W(hm-hf) 

where W = the weight of feed water lifted per pound of steam. 
hm = heat of liquid of mixture of condensed steam and 

feed water. 
hf = heat of liquid of entering feed water. 

Thermal Efficiency of Injector. The thermal efficiency of an 
injector neglecting radiaition losses is unity. All the heat ex- 
pended is restored either as work done or in heat returned to the 
boiler. 

Mechanical Efficiency. The mechanical work performed by 
the injector consists in lifting the weight of feed water and deliver- 



FLOW OF FLUIDS 



l8l 



ing it into the boiler against the internal pressure. The effi- 
ciency, considering the injector as a pump, is 

Work done 



or 



B.t.u. given up by steam to perform the work 



U 



H -h/ 
where U = { Wh -\- {W + i) h\ -^ 778 (in heat units). (117) 

Ip = pressure head corresponding to boiler gage pressure, 

in feet. 
Is = suction head in feet. 
W = pounds of water deHvered per pound of steam. 

Orifice Measurements of the flow of steam are particularly 
recommended by some engineers for ascertaining the steam con- 
sumption of the '' auxiliaries " in a power plant. This method 
commends itself particularly because of its simplicity and accu- 
racy. It is best applied by inserting a plate | inch thick with 
an orifice one inch in diameter, with square edges, at its center, 
between the two halves of a pair of flanges on the pipe through 
which the steam passes. Accurately calibrated steam gages are 
required on each side of the orifice to determine the loss of pres- 
sure. The weight of steam for the various differences of pres- 
sure may be determined by arranging the apparatus so that the 
steam passing through the orifice will be discharged into a tank 
of water placed on a platform scales. The flow through this 
orifice in pounds of dry saturated steam per hour when the dis- 
charge pressure at the orifice is 100 pounds by the gage is given 
by the following table: 



Pressure drop, 
lbs. per sq. in. 


Flow of dry steam 
per hour, lbs. 


Pressure drop, 
lbs. per sq. in. 


Flow of dry steam 
per hour, lbs. 


I 
2 

3 
4 


430 

615 

930 

1200 

1400 


5 
10 

IS 

20 


1560 
2180 
2640 
3050 







1 82 ENGINEERING THERMODYNAMICS 

Flow of Steam through Nozzles. The weight of steam dis- 
charged through any well-designed nozzle with a rounded inlet, 
similar to those illustrated in Figs. 66 and 67, depends only on 
the initial absolute pressure (Pi), if the pressure against which 
the nozzle discharges (P2) does not exceed 0.58 of the initial 
pressure. This important statement is well illustrated by the 
following example. If steam at an initial pressure (Pi) of 100 
pounds per square inch absolute is discharged from a nozzle, the 
weight of steam flowing in a given time is practically the same 
for all values of the pressure against which the steam is dis- 
charged (P2) which are equal to or less than 58 pounds per square 
inch absolute. 

If, however, the final pressure is more than 0.58 of the initial, 
the weight of steam discharged will be less, nearly in proportion 
as the difference between the initial and final pressures is re- 
duced. 

The most satisfactory and accurate formula for the " constant 
flow " condition, meaning when the final pressure is 0.58 of the 
initial pressure or less, is the following, due to Grashof ,* where w 
is the flow of steam f (initially dry saturated) in pounds per 
second, Aq is the area of the smallest section of the nozzle in 

* Grashof, Theoretische Maschinenlehre, vol. i, iii; Hiitte Taschenbuch, vol. i, 
page ^^:^. Grashof states the formula, 

■w = 0.01654 ^o-Pi,-^^> 

but |the formula given in equation (118) is accurate enough for all practical 
uses. 

t Napier's formula is very commonly used by engineers and is accurate enough 
for most calculations. It is usually stated in the form 

AoPi 

w = 5 

70 

where w, Pi, and Ao have the same significance as in Grashof s formula. The 
following formula is given by Rateau, who has done some very good theoretical 
and practical work on steam turbines, but this formula is too compHcated for 
convenient use: 

w = o.ooi AoPi [15.26 — 0.96 (log Pi + log 0.0703)]. 
Common or base 10 logarithms are to be used in this formula. 



FLOW OF FLUIDS 



183 




Fig. 66. — Example of a Well-designed Nozzle. 






DeLaval Type. 




Nozzle 

Diaphragm 



Curtis Type. 

Fig. 67. — Examples of Standard Designs of Nozzles. 



184 ENGINEERING THERMODYNAMICS 

square inches, and Pi is the initial absolute pressure of the steam 
in pounds per square inch, 

w=^^, (118) 

60 . ^ ^ 






or, in terms of the area, 

60 w 
Pi- 

These formulas are for the flow of steam initially dry and 
saturated. An illustration of their applications is given by the 
following practical example. 

Example. The area of the smallest section (^0) of a suitably 
designed nozzle is 0.54 square inch. What is the weight of the 
flow (w) of dry saturated steam per second from this nozzle when 
the initial pressure (Pi) is 135 pounds per square inch absolute 
and the discharge pressure (P2) is 15 pounds per square inch ab- 
solute? 

Here P2 is less than 0.58 Pi and Grashof 's formula is applicable, 

or, w = ^^ ) ^^^ y 

60 

0.54 X 116. 5* , , 

w = ^- = 1.049 pounds per second. 

60 

When steam passes through a series of nozzles one after the 
other as is the case in many types of turbines, the pressure is 
reduced and the steam is condensed in each nozzle so that it 
becomes wetter and wetter each time. In the low-pressure noz- 
zles of a turbine, therefore, the steam may be very wet although 
initially it was dry. Turbines are also sometimes designed to 
operate with steam which is initially wet, and this is usually the 
case when low-pressure steam turbines are operated with the 
exhaust from non-condensing reciprocating engines — a practice 
which is daily becoming 'more common. In all these cases the 
nozzle area must be corrected for the wetness of the steam. 

* The flow (w) calculated by Napier's formula for this example isw = — 

1. 041 pounds per second. 



FLOW OF FLUIDS 185 

F'or a given nozzle the weight discharged is, of course, greater 
for wet steam than for dry; but the percentage increase in the 
discharge is not nearly in proportion to the percentage of 
moisture as is often stated. The general equation for the theo- 
retic discharge (w) from a nozzle is in the form * 



w = K 



^f 



where Pi is the initial absolute pressure and vi is the specific vol- 
ume (cubic feet in a pound of steam at the pressure Pi). Now, 
neglecting the volume of the water in wet steam, which is a usual 
approximation, the volume of a pound of steam is proportional 

* The general equation for the theoretic flow is 

where the symbols w, Aq, Pi, and g are used as in equations (117) and (118). A 
is the pressure at any section of the nozzle, vi is the volume of a pound of steam at 
the pressure Pi, and ^ is a constant. The flow, w, has its maximum value when 

2 k+J. 

ik /p„\ k 



(»)-(« 



is a maximum. Differentiating and equating the first differential to zero gives 

_k 
P2 
Pi 

P2 is now the pressure at the smallest section, and writing for clearness Po for 
P2, and substituting this last equation in the formula for flow (w) above, we haxe 



{^r 



-VrJ^O (S) 



Now regardless of what the final pressure may be, the pressure (Po) at the smallest 
section of a nozzle (Ao) is always nearly 0.58 Pi for dry saturated steam. Making 
then in the last equation Po = 0.58 Pi 'and putting for k Zeuner's value of 1.135 
for dry saturated steam, we may write in general terms the form stated above, 

T Vi 

where K is another constant. See Zeuner's Theorie der Turbinen, page 268 (Ed. of 
1899). 



l86 ENGINEERING THERMODYNAMICS 

to the quality (xi). For wet steam the equation above becomes 
then 



w = K ^^ 

xiz;i 






The equation shows, therefore, that the flow of wet steam is 
inversely proportional to the square root of the quality (xi). 
Grashof s equations can be stated then more generally as 

. 60 W Vxi f . 

^0 = -^^1 (121) 

These equations become, of course, the same as (118) and 
(119) for the case where 0:1 = 1. 

Flow of Steam when the Final Pressure is more than 0.58 of 
the Initial Pressure. For this case the discharge depends upon 
the final pressure as well as upon the initial. No satisfactory 
formula can be given in simple terms, and the flow is most easily 
calculated with the aid of the curve in Fig. 68 due to Rateau. 
This curve is used by determining first the ratio of the final to 

the initial pressure — ^, and reading from the curve the corre- 
P\ 

sponding coefficient showing the ratio of the required discharge 
to that calculated for the given conditions by either of the equa- 
tions (118) or (120). The coefficient from the curve times the 
flow calculated from equations (118) or (120) is the required re- 
sult. Obviously the discharge for this condition is always less 
than the discharge when the final pressure is equal to or less 
than 0.58 of the initial. 

Length for Nozzles. Probably the best designers make the 
length of the nozzle depend only on the initial pressure. In 
other words, the length of' a nozzle for 150 pounds per square 
inch initial pressure is usually made the same for a given type 
regardless of the final pressure. And if it happens that there is 
crowding for space, one or more of the nozzles is sometimes 
made a little shorter than the others. 



FLOW OF FLUIDS 



187 



Designers of De Laval nozzles follow practically the same 
^' elastic " method. The divergence of the walls of non-con- 
densing nozzles is about 3 degrees from the axis of the nozzle, 
and condensing nozzles for high vacuums may have a divergence 



1.0 










































: 










"- 




































^ 


— 


■"" 


oT 












.9 




























-^ 














^ 


































^ 


^ 
















Q. 












.8 




















y 


^ 


















































[/ 


















































/ 


/' 


















































/ 


/ 




















































/ 


/ 












































ft S * 








/ 


/ 






















































/ 




















































/ 


















































II .4 
11-3 






/ 




















































/ 






















































/ 




















































1 






















































-2 


1 






















































I 






















































1 
















- 








































'. 














































































































.0 








• 







E{ 


iftc 


of 


Fu 


lal 


to 


Ini 


tia 


7 
LP 


resj 


3ur 


' P 


2 
1 


) 








" 


i 





Fig. 68. — Coefficients of the Discharge of Steam when the Final Pressure is 
Greater than 0.58 of the Initial Pressure. 



of as much as 6 degrees * for the normal rated pressures of the 
turbine. 

The author has used successfully the following empirical 
formula to determine. a suitable length, L, of the nozzle between 
the throat and the mouth (in inches) : 



where ^0 is the area at the throat in square inches. 



(122) 



* According to Dr. O. Recke, if the total divergence of a nozzle is more than 
6 degrees, eddies will hegin to, form in the jet. There is no doubt that a too rapid 
divergence produces a velocity loss. 

t Moyer's Steam Turbines (2d Edition), pages 45-48. 



i88 



ENGINEERING THERMODYNAMICS 



Efficiency of Nozzles. Recent experience with nozzles of this 
type does not bear out this statement, except in the case prob- 
ably of square or rectangular nozzles with no rounding at the 
edges. An efficiency of 97 per cent is not unusual for properly 
designed square and rectangular shaped nozzles without any 
*' square " edges; and circular nozzles have certainly never 
given 99 per cent efficiency. 

Under- and Over-expansion. The best efficiency of a nozzle is 
obtained when the expansion required is that for which the nozzle 
was designed, or when the expansion ratio for the condition of 
the steam corresponds with the ratio of the areas of the mouth 
and throat of the nozzle. A little under-expansion is far better, 



10 






"9 b A 



25 20 15 10 5 
percentage Nozzle is too Small' 
aX ^outli (Under Expansion) 



\- HH -W 


/ 


/ 


-I- ^ V 


/ 


-\- 4- ■ TTZ 


X 


A 


4^ Z_ 




^^^ ^^ 



5 10 15 20 25 30 

Percentage Nozzle is too Large 

at Mouth (Over Elzpansion) 



Fig. 69. — Curve of Nozzle Velocity Loss. 

however, than the same amount of over-expansion, meaning 
that a nozzle that is too small for the required expansion is more 
efficient than one that is correspondingly too large.* Fig. 69 
shows a curve representing average values of nozzle loss used by 
various American and European manufacturers \ to determine 

* It is a very good method, and one often adopted, to design nozzles so that 
at the rated capacity the nozzles under-expand at least 10 per cent, and maybe 
20 per cent. The loss for these conditions is insignificant, and the nozzles can be 
run for a large overload (with increased pressures) in nearly all types without 
immediately reducing the efficiency very much. 

t C. P. Steinmetz, Proc. Am. Soc. Mech. Engineers, May, 1908, page 628; J. 
A. Moyer, Steam Turbines, page 50. 



FLOW OF FLUIDS 189 

discharge velocities from nozzles under the conditions of under- 
or over-expansion. 

Non-expanding Nozzles. All tiie nozzles of Rateau turbines 
and usually also those of the low-pressure stages of Curtis tur- 
bines are made non-expanding; meaning, that they have the 
same area at the throat as at the mouth. For such conditions 
it has been suggested that instead of a series of separate nozzles 
in a row a single long nozzle might be used of which the sides 
were arcs of circles corresponding to the inside and outside pitch 
diameters of the blades. Advantages would be secured both on 
account of cheapness of construction and because a large amount 
of friction against the sides of nozzles would be eliminated by 
omitting a number of nozzle walls. Such a construction has not 
proved desirable, because by this method no well-formed jets 
are secured and the loss from eddies is excessive. The general 
statement may be made that the throat of a well-designed nozzle 
should have a nearly symmetrical shape, as for example a circle, 
a square, etc., rather than such shapes as ellipses and long rec- 
tangles. The shape of the mouth is not important. In Curtis 
turbines an approximately rectangular mouth is used because 
the nozzles are placed close together (usually in a nozzle plate 
like Fig. 67) in order to produce a continuous band of steam; 
and, of course, by using a section that is rectangular rather than 
circular or elliptical, a band of steam of more nearly uniform 
velocity and density is secured. 

Fig. 70 shows a number of designs of non-expanding nozzles 
used by Professor Rateau. The length of such nozzles beyond 
the throat is practically negligible. Curtis non-expanding noz- 
zles are usually made the same length as if expanding and the 
length is determined by the throat area. 

Materials for Nozzles. Nozzles for saturated or slightly super- 
heated steam are usually made of bronze. Gun metal, zinc 
alloys, and delta metal are also frequently used. All these 
metals have unusual resistance for erosion or corrosion from_ the 
use of wet steam. Because of this property as well as for the 
reason that they are easily worked with hand tools * they are 

* Nozzles of irregular shapes are usually filed by hand to the exact size. 



1 90 ENGINEERING THERMODYNAMICS 

very suitable materials for the manufacture of steam turbine 
nozzles. Superheated steam, however, rapidly erodes all these 
alloys and also greatly reduces the tensile strength. For nozzles 
to be used with highly superheated steam, cast iron is generally 
used, and except that it corrodes so readily is a very satisfactory 




Non-expanding Nozzles. 



material. Commercial copper (about 98 per cent) is said to 
have been used with a fair degree of success with high super- 
heats; but for such conditions its tensile strength is very low. 
Steel and cupro-nickel (8 Cu + 2 Ni) are also suitable materials, 
and the latter has the advantage of being practically non-cor- 
rodible. 

The most important part of the design of a nozzle is the deter- 
mination of the areas of the various sections — especially the 
smallest section, if the nozzle is of an expanding or diverging 
type. In order to calculate the areas of nozzles we must know 
how to determine the quantity of steam (flow) per unit of 
time passing through a unit area. It is very essential that the 
nozzle is well rounded on the *' entrance " side and that sharp 
edges along the path of the steam are avoided. Otherwise it 
is not important whether the shape of the section is circular, 
elliptical, or rectangular with rounded corners. 

Whether the nozzle section is throughout circular, square, or 
rectangular (if these last sections have rounded corners) the 
efficiency as measured by the velocity will be about 96 to 97 per 
cent, corresponding to an equivalent energy efficiency of 92 to 94 
per cent. Speaking commercially, therefore, it does not seem to 



FLOW OF FLUIDS 191 

be worth while to spend a great deal of time in the shops to make 
nozzles very exactly to some difficult shape. Simpler and more 
rapid methods of nozzle construction should be introduced. In 
some shops the time of one man for two days is required for the 
hand labor alone on a single nozzle. 



APPENDIX 



NAPERIAN LOGARITHMS 

c =2.7182818 log e = 0.4342945 = M 



1.0 

1.1 
1.2 
1.3 

1.4 
1.5 
1.6 

1.7 
1.8 
1.9 

2.0 

2.1 
2.2 
2.3 

2.4 
2.5 
2.6 

2.7 
2.8 
2.9 

3.0 

3.1 
3.2 
3.3 

3.4 
3.5 
3.6 

3.7 
3.8 
3.9 

4.0 

4.1 
4.2 
4.3 

4.4 
4.5 
4.6 

4.7 
4.8 
4.9 

5.0 

5.1 
5.2 
5.3 

5.4 
5.5 
5.6 



0. 0000 

0.09531 

0.1823 

0.2624 

0.3365 
0.4055 
0.4700 

0.5306 
0.5878 
0.6418 

0.6931 

0.7419 
0.7884 
0.8329 

0. 8755 
0.9163 
0.9555 

0.9933 
1.0296 
1.0647 

1.0986 

1.1314 
1.1632 
1.1939 

1.2238 
1.2528 
1.2809 

1.3083 
1.3350 
1.3610 

1.3863 

1.4110 
1.4351 
1.4586 

1.4816 
1.5041 
1.5261 

1.5476 
1.5686 
1.5892 

1.6094 

1.6292 
1.6487 
1.6677 

1.6864 
1.7047 
1.7228 



0.00995 

0. 1044 
0. 1906 
0.2700 

0.3436 
0.4121 
0.4762 

0.5365 
0.5933 
0.6471 

0.6981 

0.7467 
0.7930 
0.8372 

0.8796 
0.9203 
0.9594 

0.9969 
1.0332 
1.0682 



1.1019 

1.1346 
1.1663 
1.1969 

1.2267 
1.2556 
1.2837 

1.3110 
1.3376 
1.3635 

1.3888 

1.4134 
1.4375 
1.4609 

1.4839 
1.5063 
1.5282 



5497 
5707 
5913 



1.6114 

1.6312 

1.6506 

6696 



6882 

7066 

1.7246 



0.01980 

0.1133 
0.1988 
0.2776 

0.3507 
0.4187 
0. 4824 

0.5423 
0. 5988 
0.6523 

0.7031 

0.7514 
0.7975 
0.8416 

0.8838 
0.9243 
0.9632 

1.0006 
1.0367 
1.0716 



1.1053 

1.1378 
1.1694 
1.2000 

1.2296 
1.2585 
1.2865 

1.3137 
1.3403 
1.3661 

1.3913 

1.4159 
1.4398 
1.4633 

1.4861 
1.5085 
1.5304 

1.5518 
1.5728 
1.5933 

1.6134 

1.6332 
1.6525 
1.6715 

1.6901 

1.7884 
1.7263 



0.02956 

0. 1222 
0.2070 
0.2852 

0.3577 
0.4253 
0. 4886 

0.5481 
0. 6043 
0.6575 

0. 7080 

0.7561 
0.8020 
0. 8459 

0.8879 
0.9282 
0.9670 

1.0043 
1.0403 
1.0750 



4 



1.1086 



1.1410 
1.1725 
1.2030 



1.2326 
1.2613 
1.2892 

1.3164 
1.3429 
1.3686 

1.3938 

1.4183 
1.4422 
1.4656 

1.4884 
1.5107 
1.5326 

1.5539 
1.5748 
1.5953 

1.6154 

1.6351 
1.6544 
1.6734 

L6919 
1.7102 
1.7281 



0.03922 

0.1310 
0.2151 
0.2927 

0.3646 
0.4318 
0.4947 

0.5539 
0.6098 
0.6627 

0.7129 

0. 7608 
0. 8065 
0. 8502 

0.8920 
0.9322 
0.9708 

1.0080 
1.0438 
1.0784 

1.1119 

1.1442 
1.1756 
1.2060 

1.2355 
1.2641 
1.2920 

1.3191 
1.3455 
1.3712 

1.3962 

1.4207 
1.4446 
1.4679 



1.4907 
1.5129 
1.5347 

1.5560 
1.5769 
1.5974 

1.6174 

1.6371 
1.6563 
1.6752 

1.6938 
1.7120 
1.7299 



6 



0.04879 0.05827 0.06766 



0.1398 
0.2231 
0.3001 

0.3716 
0.4382 
0.5008 

0.5596 
0.6152 
0. 6678 

0.7178 

0.7655 
0.8109 
0.8544 

0. 8961 
0.9361 
0.9746 

1.0116 
1.0473 
1.0818 

1.1151 

1.1474 
1.1787 
1.2090 

1.2384 
1.2669 
1.2947 

1.3218 
1.3481 
1.3737 

1.3987 

1.4231 
1.4469 
1.4702 



1.4929 
1.5151 
1.5369 

1.5581 
1.5790 
1.5994 

1.6194 

1.8390 
1.6582 
1.6771 

1.6956 
1.7138 
1.7317 



0. 1484 
0.2311 
0.3075 

0.3784 
0.4447 
0.5068 

0. 5653 
0.6206 
0.6729 

0.7227 

0.7701 
0.8154 
0. 8587 

0.9002 
0.9400 
0.9783 

1.0152 
1.0508 
1.0852 

1.1184 

1.1506 
1.1817 
1.2119 

1.2413 
1.2698 
1.2975 

1.3244 
1.3507 
1.3762 

1.4012 

1.4255 
1.4493 
1.4725 

1.4951 

.5173 

1.5390 

1.5602 
1.5810 
1.6014 

1.6214 

6409 
6601 
6790 

6974 
7156 
7334 



0. 07696 



0.1570 0.1655 
0.2390 0.2469 
0.3148 0.3221 



0.3853 
0.4511 
0.5128 

0.5710 
0.6259 
0. 6780 

0.7275 

0. 7747 
0.8198 
0. 8629 

0.9042 
0.9439 
0.9821 

1.0188 
1.0543 
1.0886 

1.1217 

1.1537 
1.1848 
1.2149 

1.2442 
1.2726 
1.3002 

1.3271 
1.3533 
1.3788 

1.4036 

1.4279 
1.4516 
1.4748 

1.4974 
1.5195 
1.5412 

1.5623 
1.5831 
1.6034 

1.6233 

1.6429 

6620 

1.6808 



6993 
7174 
7352 



0.3920 
0.4574 
0.5188 

0. 5766 
0.6313 
0.6831 

0. 7324 

0.7793 
0. 8242 
0.8671 

0.9083 
0.9478 
0.9858 

1.0225 
1.0578 
1.0919 

1.1249 

1.1569 
1.1878 
1.2179 

1.2470 
1.2754 
1.3029 

1.3297 
1.3558 
1.3813 

1.4061 

1.4303 
1.4540 
1.4770 

1.4996 
1.5217 
1.5433 

1.5644 
1.5851 
1.6054 

1.6253 

1.6448 
1.6639 
1.6827 

1.7011 
1.7192 
1.7370 



0.08618 

0.1739 
0.2546 
0.3293 

0.3988 
0.4637 
0.5247 

0.5822 
0.6366 
0.6881 

0.7372 

0. 7839 
0.8286 
0.8713 

0.9123 
0.9517 
0.9895 

1.0260 
1.0613 
1.0953 

1.1282 

1.1600 
1.1909 
1.2208 

1.2499 
1.2782 
1.3056 

1.3324 

3584 

1.3838 

1.4085 

1.4327 

1.4563 

4793 

1.5019 
1.5239 
1.5454 

1.5665 

1.5872 
1.6074 

1.6273 



6467 

6658 

1.6845 



7029 
7210 
7387 



(195) 



NAPERIAN LOGARITHMS. 








1 


2 


3 


4 


5 


6 


7 


8 


9 


6.7 
5.8 
6.9 


1.7405 
1.7579 
1.7750 


1.7422 
1.7596 
1.7766 


1.7440 
1.7613 
1.7783 


1.7457 
1.7630 
1.7800 


1.7475 
1.7647 
1.7817 


1.7492 
1.7664 
1.7834 


1.7509 
1.7681 
1.7851 


1.7527 
1.7699 
1.7867 


1.7544 
1.7716 
1.7884 


1.7561 
1.7733 
L7901 


6.0 


1.7918 


1.7934 


1.7951 


1.7967 


1.7984 


1.8001 


1.8017 


1.8034 


1.8050 


1.8066 


6.1 
6.2 
6.3 


1.8083 
1.8245 
1.8405 


1.8099 
1.8262 
1.8421 


1.8116 

1.8278 
1.8437 


1.8132 
1.8294 
1.8453 


1.8148 
1.8310 
1.8469 


1.8165 
1.8326 
1.8485 


1.8181 
1.8342 
1.8500 


1.8197 
1.8358 
1.8516 


1.8213 
1.8374 
1.8532 


1.8229 
1.8390 
1.8547 


6.4 
6.5 
6.6 


1.8563 
1.8718 
1.8871 


1.8579 
1.8733 
1.8886 


1.8594 
1.8749 
1.8901 


1.8610 

1.8764 
1.8916 


1.8625 
1.8779 
1.8931 


1.8641 
1.8795 
1.8946 


1.8656 
1.8810 
1.8961 


1.8672 
1.8825 
1.8976 


1.8687 
1.8840 
1.8991 


1.8703 
1.8856 
1.9006 


6.7 
6.8 
6.9 


1.9021 
1.9169 
1.9315 


1.9036 
1.9184 
1.9330 


1.9051 
1.9199 
1.9344 


1.9066 
1.9213 
1.9359 


1.9081 
1.9228 
1.9373 


1.9095 
1.9242 
1.9387 


1.9110 
1.9257 
1.9402 


1.9125 
1.9272 
1.9416 


1.9140 
1.9286 
1.9430 


1.9155 
1.9301 
1.9445 


7.0 


1.&459 


1.9473 


1.9488 


1.9502 


1.9516 


1.9530 


1.9544 


1.9559 


1.9573 


1.9587 


7.1 
7.2 
7.3 


1.9601 
1.9741 
1.9879 


1.9615 
1.9755 
1.9892 


1.9629 
1.9769 
1.9906 


1.9643 
1.9782 
1.9920 


1.9657 
1.9796 
1.9933 


1.9671 
1.9810 
1.9947 


1.9685 
1.9824 
1.9961 


1.9699 
1.9838 
1.9974 


1.9713 
1.9851 
1.9988 


1.9727 
1.9865 
2.0001 


7.4 
7.5 
7.6 


2.0015 
2.0149 
2.0281 


2.0028 
2.0162 
2.0295 


2.0042 
2.0176 
2.0308 


2. 0055 
2.0189 
2.0321 


2.0069 
2.0202 
2.0334 


2. 0082 
2.0215 
2.0347 


2.0096 
2. 0229 
2.0360 


2.0109 
2.0242 
2.0373 


2.0122 
2.0255 
2.0386 


2.0136 
2.0268 
2.0399 


7.7 
7.8 
7.9 


2.0412 
2.0541 
2.0668 


2.0425 
2.0554 
2.0681 


2.0438 
2.0567 
2.0694 


2.0451 
2.0580 
2.0707 


2.0464 
2.0592 
2.0719 


2.0477 
2.0605 
2.0732 


2.0490 
2.0618 
2.0744 


2.0503 
2.0631 
2.0757 


2.0516 
2.0643 
2.0769 


2.0528 
2.0656 
2. 0782 


8.0 


2.0794 


2.0807 


2.0819 


2.0832 


2.0844 


2.0857 


2.0869 


2.0881 


2.0894 


2.0906 


8.1 
8.2 
8.3 


2.0919 
2.1041 
2.1163 


2.0931 
2.1054 
2.1175 


2.0943 
2.1066 
2.1187 


2.0956 
2.1078 
2.1199 


2.0968 
2.1090 
2.1211 


2.0980 
2.1102 
2.1223 


2.0992 
2.1114 
2.1235 


2. 1005 
2.1126 
2.1247 


2.1017 
2.1138 
2.1258 


2. 1029 
2.1150 
2. 1270 


8.4 
8.5 
8.6 


2. 1282 
2.1401 
2.1518 


2. 1294 
2.1412 
2. 1529 


2.1306 
2. 1424 
2.1541 


2.1318 
2.1436 
2. 1552 


2.1330 
2. 1448 
2. 1564 


2.1342 
2.1459 
2.1576 


2.1353 
2.1471 
2.1587 


2.1365 
2.1483 
2.1599 


2.1377 
2.1494 
2.1610 


2. 1389 
2.1506 
2.1622 


8.7 
8.8 
8.9 


2.1633 
2. 1748 
2. 1861 


2. 1645 
2.1759 
2.1872 


2.1656 
2.1770 
2. 1883 


2.1668 
2.1782 
2. 1894 


2. 1679 
2.1793 
2. 1905 


2.1691 

2.1804 
2.1917 


2. 1702 

2.1815 
2.1928 


2.1713 
2.1827 
2.1939 


2.1725 
2.1838 
2. 1950 


2.1736 
2. 1849 
2.1961 


9.0 


2. 1972 


2.1983 


2. 1994 


2.2006 


2.2017 


2. 2028 


2. 2039 


2.2050 


2.2061 


2.2072 


9.1 
9.2 
9.3 


2.2083 
2.2192 
2. 2300 


2.2094 
2.2203 
2. 2311 


2.2105 

2.2214 
2. 2322 


2.2116 
2.2225 
2.2332 


2.2127 
2.2235 
2. 2343 


2.2138 
2.2246 
2.2354 


2.2148 
2.2257 
2.2364 


2.2159 
2.2268 
2.2375 


2.2170 
2.2279 
2. 2386 


2.2181 
2. 2289 
2.2396 


9.4 
9.5 
9.6 


2. 2407 
2.2513 
2.2618 


2.2418 
2.2523 
2.2628 


2.2428 
2.2534 
2.2638 


2.2439- 

2.2544 

2.2649 


2. 2450 
2.2555 
2.2659 


2.2460 
2.2565 
2.2670 


2.2471 
2.2576 
2.2680 


2.2481 
2.2586 
2.2690 


2.2492 
2.2597 
2.2701 


2.2502 
2.2607 
2.2711 


9.7 
9.8 
9.9 


2.2721 
2.2824 
2. 2925 


2.2732 
2.2834 
2.2935 


2. 2742 
2. 2844 
2.2946 


2.2752 
2.2854 
2.2956 


2.2762 
2.2865 
2.2966 


2.2773 

2.2875 
2.2976 


2.2783 
2.2885 
2.2986 


2.2793 
2.2895 
2.2996 


2.2803 
2.2905 
2.3006 


2.2814 
2.2915 
2.3016 


10.0 


2.3026 





















(196) 



LOGARITHMS. 



Nat. 
Nos. 
























Proportion 


a Parts. 1 





1 


2 


3 


4 


5 


6 


7 


8 


9 






































1 2 


3 


4 


5 


6 


7 


8 9 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


4 8 


12 


17 21 


25 


29 


33 37 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


4 8 


11 


15 


19 


23 


26 


30 34 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


3 7 


10 


14 


17 21 


24 


28 31 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


3 6 


10 


13 


16 


19 


23 


26 29 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


3 6 


9 


12 


15 


18 


21 


24 27 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


3 6 


8 


11 


14 


17 


20 


22 25 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


3 5 


8 


11 


13 


16 


18 


21 24 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


2 5 


7 


10 


12 


15 


17 


20 22 


18 


2553 


2577 


2601 


'>625 


2648 


2672 


2695 


2718 


2742 


2765 


2 5 


7 


9 


12 


14 


16 


19 21 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


2 4 


7 


9 


11 


13 


16 


18 20 


20 


3010 


3032 


3054 


3075 


3096 




3118 


3139 


3160 


3181 


3201 


2 4 


6 


8 


11 


13 


15 


17 19 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


2 4 


6 


8 


10 


12 


14 


16 18 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


2 4 


6 


8 


10 


12 


14 


15 17 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


2 4 


6 


7 


9 


11 


13 


15 17 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


2 4 


5 


7 


9 


11 


12 


14 16 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


2 3 


5 


7 


9 


10 


12 


14 15 


26 


4150 


4166 4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


2 3 


5 


7 


8 


10 


11 


13 15 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


2 3 


5 


6 


8 


9 


11 


13 14 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


2 3 


5 


6 


8 


9 


11 


12 14 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


1 3 


4 


6 


7 


9 


10 


12 13 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


1 3 


4 


6 


7 


9 


10 


11 13 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


1 3 


4 


6 


7 


8 


10 


11 12 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


1 3 


4 


5 


7 


8 


9 


11 12 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 


1 3 


4 


5 


6 


8 


9 


10 12 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


1 3 


4 


5 


6 


8 


9 


10 11 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


1 2 


4 


5 


6 


7 


9 


10 11 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


1 2 


4 


5 


6 


7 


8 


10 11 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


1 2 


3 


5 


6 


7 


8 


9 10 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


1 2 


3 


5 


6 


7 


8 


9 10 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


1 2 


3 


4 


5 


7 


8 


9 10 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


1 2 


3 


4 


5 


6 


8 


9 10 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


1 2 


3 


4 


5 


6 


7 


8 9 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


1 2 


3 


4 


5 


6 


7 


8 9 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


1 2 


3 


4 


5 


6 


7 


8 9 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


1 2 


3 


4 


5 


6 


7 


8 9 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


1 2 


3 


4 


5 


6 


7 


8 9 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


1 2 


3 


4 


5 


6 


7 


7 8 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


1 2 


3 


4 


5 


5 


6 


7 8 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


1 2 


3 


4 




5 


6 


7 8 


49 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


1 2 


3 


4 




5 


6 


7 8 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


1 2 


3 


3 




5 


6 


7 8 


51 


7076 


7084 


7093 


7101 


7U0 


7118 


7126 


7135 


7143 


7152 


1 2 


3 


3 




5 


6 


7 8 


52 


7160 


7168 


7177 7185 1 


7193 


7202 


7210 


7218 


7226 


7235 


1 2 


2 


3 




5 


6 


7 7 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


1 2 


2 


3 




5 


6 


6 7 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


1 2 


^ 


3 




5 


6 


6 7 



(197) 



LOGARITHMS. 



Nat. 
Nos. 





1 


2 


3 


4 


5 


6 


7 


Q 


9 


Proportional Parts. 


12 3 


4 5 6 


7 8 9 


55 
56 
57 
58 
59 

60 

61 
62 
63 
64 


7404 

7482 
7559 
7634 
7709 

7782 
7853 
7924 
7993 
8062 


7412 
7490 
7566 
7642 
7716 

7789 
7860 
7931 
8000 
8069 


7419 
7497 
7574 
7649 
7723 

7796 
7868 
7938 
8007 
8075 


7427 
7505 
7582 
7657 
7731 

7803 

7875 
7945 
8014 
8082 


7435 
7513 
7589 
7664 
7738 

7810 
7882 
7952 
8021 
8089 


7443 
7520 
7597 
7672 
7745 

7818 
7889 
7959 
8028 
8096 


7451 

7528 
7604 
7679 
7752 

7825 
7896 
7966 
8035 
8102 


7459 
7536 
7612 
7686 
7760 

7832 
7903 
7973 
8041 
8109 


7466 
7543 
7619 
7694 
7767 

7839 
7910 
7980 
8048 
8116 


7474 
7551 
7627 
7701 

7774 

7846 
7917 
7987 
8055 
8122 


1 2 2 
1 2 2 
1 2 2 
1 1 2 
1 1 2 

1 1 2 
1 1 2 
1 1 2 
1 1 2 
1 1 2 


3 4 5 
3 4 5 
3 4 5 
3 4 4 
3 4 4 

3 4 4 
3 4 4 
3 3 4 
3 3 4 
3 3 4 


5 6 7 
5 6 7 
5 6 7 
5 6 7 
5 6 7 

5 6 6 
5 6 6 
5 6 6 
5 5 6 
5 5 6 


65 
66 
67 
68 
69 

70 
71 
72 
73 

74 


8129 
8195 
8261 
8325 
8388 

8451 
8513 
8573 
8G33 
8692 


8136 
8202 
8267 
8331 
8395 

8457 
8519 
8579 
8639 
8698 


8142 
8209 
8274 
8338 
8401 

8463 
8525 
8585 
8645 
8704 


8149 
8215 
8280 
8344 
8407 

8470 
8531 
8591 
8651 
8710 


8156 
8222 
8287 
8351 
8414 

8476 
8537 
8597 
8657 
8716 


8162 

8228 
8293 
8357 
8420 

8482 
8543 
8603 
8663 
8722 


8169 
8235 
8299 
8363 
8426 

8488 
8549 
8609 
8669 

8727 


8176 
8241 
8306 
8370 
8432 

8494 
8555 
8615 
8675 
8733 


8182 
8248 
8312 
8376 
8439 

8500 
8561 
8621 
8681 
8739 


8189 
8254 
8319 
8382 
8445 

8506 
8567 
8627 
8686 
8745 


1 1 2 
1 1 2 
1 1 2 
1 1 2 
1 1 2 

1 1 2 
1 1 2 
1 1 2 
1 1 2 

1 1 2 


3 3 4 
3 3 4 
3 3 4 
3 3 4 
2 3 4 

2 3 4 
2 3 4 
2 3 4 
2 3 4 
2 3 4 


5 5 6 
5 5 6 
5 5 6 
4 5 6 
4 5 6 

4 5 6 
4 5 5 
4 5 5 
4 5 5 
4 5 5 


75 
76 
77 
78 
79 

80 
81 
82 
83 

84 


8751 
8808 
8865 
8921 
8976 

9031 
9085 
9138 
9191 
9243 


8756 
8814 
8871 
8927 
8982 

9036 
9090 
9143 
9196 
9248 


8762 
8820 
8876 
8932 
8987 

9042 
9096 
9149 
9201 
9253 


8768 
8825 
8882 
8938 
8993 

9047 
9101 
9154 
9206 
9258 


8774 
8831 
8887 
8943 
8998 

9053 
9106 
9159 
9212 
9263 


8779 
8837 
8893 
8949 
9004 

9058 
9112 
9165 
9217 
9269 


8785 
8842 
8899 
8954 
9009 

9063 
9117 
9170 
9222 
9274 


8791 
8848 
8904 
8960 
9015 

9069 
9122 
9175 
9227 
9279 


8797 
8854 
8910 
8965 
9020 

9074 
9128 
9180 
9232 
9284 


8802 
8859 
8915 
8971 
9025 

9079 
9133 
9186 
9238 
9289 


1 1 2 
1 1 2 
1 1 2 
1 1 2 
1 1 2 

1 1 2 
1 1 2 
1 1 2 
1 1 2 
1 1 2 


2 3 3 
2 3 3 
2 3 3 
2 3 3 
2 3 3 

2 3 3 
2 3 3 
2 3 3 
2 3 3 
2 3 3 


4 5 5 
4 5 5 
4 4 5 
4 4 5 
4 4 5 

4 4 5 
4 4 5 
4 4 5 
4 4 5 
4 4 5 


85 
86 
87 
88 
89 

90 
91 
92 
93 
94 


9294 
9345 
9395 
9445 
9494 

9542 
9590 
9638 
9685 
9731 


9299 
9350 
9400 
9450 
9499 

9547 
9595 
9643 
9689 
9736 


9304 
9355 
9405 
9455 
9504 

9552 
9600 
9647 
9694 
9741 


9309 
9360 
9410 
9460 
9509 

9557 
9605 
9652 
9699 
9745 


9315 
9365 
9^_15 
9465 
9513 

9562 
9609 
9657 
9703 
9750 


9320 
9370 
9420 
9469 
9518 

9566 
9614 
9661 
9708 
9754 


9325 
9375 
9425 
9474 
9523 

9571 
9619 
9666 
9713 
9759 


9330 
9380 
9430 
9479 
9528 

9576 
9G24 
9671 
9717 
9763 


9335 
9385 
9435 
9484 
9533 

9581 
9628 
9675 
9722 
9768 


9340 
9390 
9440 
9489 
9538 

9586 
9633 
9680 

9727 
9773 


1 1 2 
1 1 2 
1 1 
1 1 
1 1 

1 1 
11 
1 1 
1 1 
1 1 


2 3 3 
2 3 3 
2 2 3 
2 2 3 
2 2 3 

2 2 3 
2 2 3 
2 2 3 
2 2 3 
2 2 3 


4 4 5 
4 4 5 
3 4 4 
3 4 4 
3 4 4 

3 4 4 
3 4 4 
3 4 4 
3 4 4 
3 4 4 


95 
96 
97 
98 
99 


9777 
9823 
9868 
9912 
9956 


9782 
9827 
9872 
9917 
9961 


9786 
9832 
9877 
9921 
9965 


9791 
9836 
9881 
9926 
9969 


9795 
9841 
9886 
9930 

9974 


9800 
9845 
9890 
9934 
9978 


9805 
9850 
9894 
9939 
9983 


9809 
9854 
9899 
9943 
9987 


9814 
9859 
9903 
9948 
9991 


9818 
9863 
9908 
9952 
9996 


1 1 
1 1 
1 1 
1 1 
1 1 


2 2 3 
2 2 3 
2 2 3 
2 2 3 
2 2 3 


3 4 4 
3 4 4 
3 4 4 
3 4 4 
3 3 4 



(198) 





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(T99) 



INDEX 



Absolute temperature, ii. 

pressure, 13. 

zero, II, 
Absorption system of refrigeration, 90. 
Adiabatic expansion, 30, 35, 39, 114. 

compression, 35, 39, 46. 

lines of steam, 108. 
Air compressor, 95-99. 

engines, 54, 56. 

thermometer, 14. 
Air, table of properties of, 23. 
Ammonia machine, 92-94. 
Available energy of steam, 120-127, 129. 

Barrel calorimeter, 82. 
Barrus' calorimeter, 78. 
Black's doctrine of latent heat, i. 
Boyle's law, 10, 14. 
British thermal unit, 4, 5. 

Calorie, 4. 

Calorimeter, steam, 75-83. 

Camot, I, 44. 

Camot's cycle, 44. 

cycle, efficiency of, 48. 

cycle, reversed, 51, 88, 

cycle principle, 49. 
Centigrade, 4, 12. 
Charles' law, 12, 14. 
Clausius, I. 

cycle, 127. 
Coefficient of performance, 93-94. 
Combined diagrams, 96, 157. 
Combination law, 14, 36. 
Compressed air, 95-99. 
Compression, isothermal, 32, 95. 

of gases, 29. 
Compressor, air, 95-99. 
Condensing calorimeters, 82. 
Conservation of energy, 3, 



Conversion of pressures, 13 (foot-note). 

of temperatures, 4, 11. 
Cross-section paper, logarithmic, 199. 
Cycle, Carnot's, 44, 

Clausius, 127. 

definition of, 44, 140. 

Rankine, 127. 

Dense air machine, 90-92. 

Density, 4, 23. 

Diagram, temperature-entropy,io5-io7. 

indicator, 30, 157. 

Mollier, 116, 117. 

total heat-entropy, 116, 117. 
Dry saturated steam, 61. 
Drying of steam by throttling, 73. 

Efficiency, air engines, 56. 

Carnot cycle, 48, 

Clausius cycle, 127, 

Ericsson engine, 56. 

mechanical, 141. 

non-expansive cycle, 144, 145. 

Rankine cycle, 127, 129, 139. 

refrigerating machine, 94, 

Stirling engine, 56, 

thermal, 6, 56. 
Energy, available, 120-127, 129. 

internal, 20, 38, 66. 

intrinsic, 20, 
Engine, Ericsson, 56. 

hot air, 56. 

Stirling, 56. 
Engineering units, 4. 
Entropy, 10 1. 

diagram, 105-107. 

o^ the evaporation, 107. 

of the liquid, 107. 
Equivalent evaporation, 8^. 
Ericsson hot-air engine, 56. 



202 



INDEX 



Evaporation, equivalent, 83. 

factor of, 83. 

internal energy of, 66. 

latent heat of, 63. 
Expansion of gases, 29. 
Expansions, adiabatic, 30, 35, 39. 

isothermal, 30, 32, 102. 
External work, 19, 67. 

work in steam formation, 61, 64. 

Factor of evaporation, 83. 
First law of thermodynamics, 3. 
Flow of air, 164-172. 

in nozzles, 163, 172. 

in orifices, 168-171, 181. 

of steam, 172-189. 
Foot-pound, 4. 
Foot, square, 4. 
Formation of steam, 60. 

Gas constant (R), 16, 23. 

thermometers, 13. 
Gas, perfect, 9. 
Gram-calorie, 4. 

h (heat of liquid), 62. 

H (total heat of steam), 65. 

Heat engine, efficiencies, 139, 144. 

of liquid, 62. 

units, 4. 
Him's analysis, 160. 
Historical, i. 
Hot-air engine, 54, 56. 
Hyperbolic logarithms, 195, 196. 

Indicator diagram, 30. 
Internal energy, 20, 38, 66, 67. 

work of evaporation, 66. 
Intrinsic energy, 20, 38, 66. 
Irreversible cycle, 50. 
Isentropic lines, 103 (f oot-n'ote) . 
Isothermal expansion, 30, 32, 46. 

compression, 32, 46. 

lines of steam, 108. 

Joule, I, 21. 
Joules' law, 21. 

Kelvin, i, 21. 



L (latent heat of steam), 6z. 
Latent heat of evaporation, 63. 
Laws of perfect gases, 10, 12, 14, 30, 37, 
39. 48. 

of thermodynamics, 3. 
Logarithmic cross-section paper, 199. 
Logarithms, natural, 34, 195, 196. 

"common" or ordinary, 34, 197, 198. 
Low temperature researches, 24. 

M (mass), 17. 

Mass, 17. 

Mean specific heat, 72. 

Mechanical efficiency, 141. 

equivalent of heat, 5. 
Moisture in steam, 69, 74. 
MoUier diagram, 116, 117. 
Moyer's formula for nozzle losses, 188. 

Naperian logarithms, 195, 196. 
Natural logarithms, 34, 195, 196. 
Newcomen, i. 
Non-expansive cycle, 113. 
Non-reversible cycle, 50, 51. 
Nozzle, flow through, 123. 
Nozzle losses, 175, 188. 

Onnes, 24. 

Orifice, flow through, 163, 172. 

Perfect gas, 9, 39. 
Perfection of heat engine, 51. 
Porous plug experiment, 21. 
Pound, 4. 

Pressure-temperature relation, 12, 39, 
62. 

units, 4. 
Properties of gases, 23. 

of steam, 61. 

9,63. 

QuaUty of steam, 69, 77, 114. 

R (thermodynamic or "gas" constant), 

16, 17, 21, 23. 
r, 64. 
Rankine, i. 

cycle, 127, 129, 139. 



INDEX 



203 



Ratio of expansion (;-), 35, 41. 
Ratio of specific heats, 23. 
Refrigerating machines, 44, 88-94. 
Refrigeration, appHcations of, 88-94. 
Regenerator, 55. 
Regnault, i, 62. 
Reversed Carnot cycle, 51. 
Reversibility, 49. 
Reversible cycle, 49. 

Saturated steam, 61. 
Saturation curve, 138, 158. 

line of steam, 159. 
Scales, thermometric, 4. 
Second law of thermodynamics, 3, 51, 

53- 
Separating calorimeter, 80. 
Small calorie, 5 (foot-note). 
Specific heat, 18. 

heat at constant volume, 19. 

heat, difference of, 19. 

heat, instantaneous values of, 73. 

heat, mean value of, 72. 

heat, ratio of, 23. 

heat, true, 72, 73. 

heat of gases, 23. 

heat of superheated steam, 72, 73. 

heat of water, 5. 

volume of gases, 4, 23. 

volume of saturated steam, 64. 

volume of superheated steam, 72. 
Steam, dry, 61. 

engines, efficiencies of, 144, 147. 

formation of, 60. 

saturated, 61. 

superheated, 61, 62. 

tables, 60. 

total heat of, 65, 

turbine, 123, 148. 

wet, 68. 
Stirling's engine, 56. 
Superheated steam, 61, 62, 69. 
Superheating calorimeter, 75. 



Tables, steam, 60. 
Temperature, absolute, 11. 

-entropy diagrams, 105-107. 

measurements, low, 24. 
Thermal efficiency, 6, 56, 141. 

unit, British, 4, 5. 
Thermodynamics, definition of, i. 
Thermometer, air, 14. 

gas, 13. 
Thermometric scales, 4. 
Throtthng calorimeter, 75, 78. 
Total heat-entropy diagram, ji6, 117. 

heat of saturated steam, 65. 

heat of superheated steam, 70. 

internal energy of steam, 66. 
Turbine, steam, 123. 

Unit of heat, i, 4, 5. 
Units, engineering, 4. 

Vaporization (see evaporation). 
Vapors, examples of, 90. 
Velocity, 123. 
Volume, specific, 4, 23. 

Water, specific heat of, 5. 

weight of, 13 (foot-note). 
Watt, I. 

Weight, units of, 4, 13 (foot-note). 
Wet steam, 68. 
Wetness of steam, 69, 77. 
Wire-drawing, 73. 
Work, external, 19, 67. 

of adiabatic expansion, 58. 

of Carnot cycle, 48. 

of Clausius cycle, 128. 

of isothermal expansion, 34. 

of Rankine cycle, 128. 
Working substance, 6, 7. 

X (quality of steam), 69, 77. 

Zero, absolute, 11. 



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